Work and energy

Class 11 Physics, unit 5, book pages 77–87

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Work is the way a force transfers energy to a body, and this unit shows how to calculate it, including for a force that changes along the path. It separates conservative fields (gravity), where work does not depend on the path, from non-conservative ones (friction). It then defines kinetic energy from the equations of motion and links work and energy through the work-energy theorem, including motion against a resistive force and the efficiency of real machines.

Learning outcomes (4)

Derive the formula for kinetic energy [using the equations of motion].
Deduce the work done from force-displacement graph.
Differentiate between conservative and non-conservative forces.
Utilize the work-energy theorem in a resistive medium to solve problems.

5.1WorkOften examined

In the sport of weightlifting the task is to pick up a very large mass, lift it over the head and hold it there at rest for a moment. This is an example of doing work by lifting and lowering a mass. It needs energy from the body's metabolic processes, specifically from the breakdown of a molecule called adenosine triphosphate (ATP), and stored glycogen in muscle tissue can also be broken down into glucose to provide energy.

The work done on an object is the scalar (or dot) product of the force $F$ and the displacement $d$ .

Work: The scalar (dot) product of the force $F$ acting on an object and the displacement $d$ of the object: $W = F \cdot d = F d cos θ$ .

W = F \cdot d or W = F d cos θ

5.1

In Eq. (5.1), $F$ is the magnitude of the force, $d$ is the magnitude of the displacement and $θ$ is the angle between the force and the displacement. In Figure 5.1 the force $F$ is resolved into two components, $F cos θ$ and $F sin θ$ . The component $F cos θ$ lies along the displacement and is the effective component. The component $F sin θ$ is perpendicular to the displacement and plays no role in doing work.

Figure 5.1: work done on an object when a force

F

produces displacement

d

Units of work

The SI unit of work is the joule (J), named in honour of the 19th-century English physicist James Prescott Joule. Eq. (5.1) shows that the unit of work is the unit of force multiplied by the unit of distance. In SI the unit of force is the newton and the unit of distance is the metre, so one joule is one newton-metre.

1 J = 1 N m

Dependence of work

The work done depends upon the force $F$ , the displacement $d$ and the angle $θ$ between them, as shown in Figure 5.2. The sign of the work depends on $θ$ through $cos θ$ .

Figure 5.2: (a) positive work, (b) negative work, (c) zero work.

Positive work. When the force has a component in the same direction as the displacement ( $0^{\circ} \leq θ < 9 0^{\circ}$ ), $cos θ$ in Eq. (5.1) is positive and the work $W$ is positive.
Negative work. When the force has a component opposite to the displacement ( $9 0^{\circ} < θ \leq 18 0^{\circ}$ ), $cos θ$ is negative and the work done by the force is negative.
Zero work. When the force is perpendicular to the displacement ( $θ = 9 0^{\circ}$ ), the work done by the force is zero.

Sign of the work done by a constant force.
Angle $θ$	$cos θ$	Work $W = F d cos θ$	Component $F cos θ$
$0^{\circ} \leq θ < 9 0^{\circ}$	positive	positive	along the displacement
$θ = 9 0^{\circ}$	$0$	zero	none ( $F cos θ = 0$ )
$9 0^{\circ} < θ \leq 18 0^{\circ}$	negative	negative	opposite to the displacement

Example 5.1

Mehwish is dragging a suitcase at an airport and pulls it through a distance of

4.90 m

along level ground. She is applying a constant force of

16.8 N

3 0^{\circ}

with the horizontal. A

0.170 N

friction force opposes the suitcase's motion. Find the work done by (a) Mehwish, (b) the frictional force and (c) the net work done by all the forces acting on the suitcase.

Given

$F_{M} = 16.8 N$ force by Mehwish
$θ_{M} = 3 0^{\circ}$
$F_{f} = 0.170 N$ friction
$θ_{f} = 18 0^{\circ}$
$d = 4.90 m$

To find

$W_{M}$
$W_{f}$
$W_{n e t}$

Example 5.1: the suitcase is pulled with

F_{M} = 16.8 N

3 0^{\circ}

through

d = 4.90 m

, against friction

F_{f} = 0.170 N

(a) Work done by Mehwish. Use Eq. (5.1) with her force and the angle $3 0^{\circ}$ .
$W_{M} = F_{M} d cos θ_{M} = 16.8 \times 4.90 \times cos 3 0^{\circ} = 71.3 N m$
Write the result in joules.
$W_{M} = 71.3 J$
(b) Work done by friction. Friction points opposite to the displacement, so $θ_{f} = 18 0^{\circ}$ and $cos 18 0^{\circ} = - 1$ .
$W_{f} = F_{f} d cos θ_{f} = 0.170 \times 4.90 \times cos 18 0^{\circ} = - 0.833 J$
(c) The weight of the suitcase and the normal reaction are perpendicular to the motion, so they have no effective component and do no work. The net work is the sum of the work done by Mehwish and by friction.
$W_{n e t} = W_{M} + W_{f} = 71.3 + (- 0.833) = 70.467 J \approx 70.5 J$

W_{M} = 71.3 J, W_{f} = - 0.833 J, W_{n e t} = 70.5 J

Assignment 5.1

A box having

40 kg

mass is dragged on a frictionless inclined surface to a height of

8 m

. If the inclined plane makes an angle of

2 0^{\circ}

with the Earth, find the work done against gravity.

Given

$m = 40 kg$
$h = 8 m$
$α = 2 0^{\circ}$ angle of the incline
$g = 9.8 m s^{- 2}$

To find

$W$ against gravity

The box moves up the slope, so the length of the slope travelled is the displacement.
$d = \frac{h}{sin 2 0 ^{\circ}} = \frac{8}{0.342} = 23.4 m$
Gravity acts down the slope with the component $m g sin α$ . The box is dragged without friction, so the force needed (and the force gravity opposes) is this component.
$F = m g sin 2 0^{\circ} = 40 \times 9.8 \times 0.342 = 134 N$
The force and the displacement are in the same direction ( $θ = 0^{\circ}$ ) for the dragging force, so the work done against gravity is
$W = F d = 134 \times 23.4 = 3.14 \times 1 0^{3} J$
Check with the vertical height. Only the height gained matters, because gravity is conservative (Section 5.2).
$W = m g h = 40 \times 9.8 \times 8 = 3136 J$

W = m g h = 3136 J \approx 3.14 kJ

Worked example: work done by each force on a pulled box

20 kg

box is pulled

6.0 m

along a level floor by a rope. The rope pulls with

50 N

6 0^{\circ}

above the horizontal and the floor exerts a friction force of

10 N

. Find the work done on the box by (a) the pull, (b) friction, (c) the weight, (d) the normal reaction and (e) all the forces together.

Given

$F = 50 N at 6 0^{\circ}$
$f = 10 N$
$d = 6.0 m$
$m = 20 kg$

To find

(a) to (d) work done by each force
(e) net work

(a) Only the part of the pull along the displacement works, so use $W = F d cos θ$ with $θ = 6 0^{\circ}$ between the rope and the floor.
$W_{F} = 50 \times 6.0 \times cos 6 0^{\circ} = 50 \times 6.0 \times 0.5 = 150 J$
(b) Friction points opposite to the displacement, so $θ = 18 0^{\circ}$ and $cos 18 0^{\circ} = - 1$ . The work is negative.
$W_{f} = 10 \times 6.0 \times (- 1) = - 60 J$
(c) The weight acts straight down and the box moves horizontally, so $θ = 9 0^{\circ}$ and $cos 9 0^{\circ} = 0$ .
$W_{w} = 0$
(d) The normal reaction acts straight up, also at $9 0^{\circ}$ to the displacement, so it does no work either.
$W_{N} = 0$
(e) Work is a scalar, so add the four values with their signs.
$W_{n e t} = 150 - 60 + 0 + 0 = 90 J$

Figure: the four forces on the box. Only the pull

F

(accent) has a component along

d

, so it alone does positive work.

W_{F} = 150 J, W_{f} = - 60 J, W_{w} = W_{N} = 0, W_{n e t} = 90 J

5.1Work done from force-displacement graphOften examined

The area under the force-displacement graph gives the work done.

A. Work done by a constant force

Figure 5.3 shows the graph of a constant force producing a net displacement. The shaded area of the rectangle represents the work done by the constant force: height $F$ times width $d$ is $F d$ .

Work done by a constant force

Displacement (m)

Force (N)

W

F

Figure 5.3: work done by a constant force is the area of the rectangle.

B. Work done by a variable force

Figure 5.4 shows a force that increases with displacement. The shaded area of the triangle represents the work done by this increasing force. In many situations in daily life the force varies in many different ways, so the area under the curve takes different shapes.

Work done by a variable force

Displacement (m)

Force (N)

W

F

Figure 5.4: work done by a variable (increasing) force is the area of the triangle.

For example, when we stretch a spring, the more we stretch it the harder we have to pull, so the force we exert is not constant as the spring is stretched. When a rocket moves away from the Earth, the work is done against the force of gravity, which varies as the inverse of the square of the distance from the centre of the Earth.

Derivation Work done by a variable force as an area

Aim. Show that the work done by a variable force is equal to the area under the $F cos θ$ against $d$ curve.

Assumptions.

A graph of $F cos θ$ against $d$ is drawn for the motion (Figure 5.5).
The displacement can be divided into small displacements $Δ d_{1}, Δ d_{2}, Δ d_{3}, \dots, Δ d_{n}$ , small enough for the effective force to be taken as constant over each one.

Figure 5.5: graphical analysis of work done by a variable force. Left: the displacement is divided into strips

Δ d

. Right: the work is the area under the curve.

Divide the whole displacement into $n$ small displacements. Over each one the effective component of the force is nearly constant: $F_{1} cos θ_{1}, F_{2} cos θ_{2}, F_{3} cos θ_{3}, \dots, F_{n} cos θ_{n}$ .
For each strip the work is the constant-force result $F d cos θ$ , which is the area of that strip (height times width).
$W_{1} W_{2} W_{3} W_{n} = F_{1} cos θ_{1} Δ d_{1} = F_{2} cos θ_{2} Δ d_{2} = F_{3} cos θ_{3} Δ d_{3} ⋮ = F_{n} cos θ_{n} Δ d_{n}$
The total work done is the sum of all the individual works.
$W_{T o t a l} = W_{1} + W_{2} + W_{3} + \dots + W_{n}$
Put in the work for each strip.
$W_{T o t a l} = F_{1} cos θ_{1} Δ d_{1} + F_{2} cos θ_{2} Δ d_{2} + F_{3} cos θ_{3} Δ d_{3} + \dots + F_{n} cos θ_{n} Δ d_{n} (5.2)$
Write the sum in compact form with the summation sign.
$W_{T o t a l} = i = 1 \sum n F_{i} cos θ_{i} Δ d_{i} (5.3)$

Result

W_{T o t a l} = i = 1 \sum n F_{i} cos θ_{i} Δ d_{i} = area under the F cos θ against d curve

The sum of the strip areas is the area under the curve. The thinner the strips, the better the sum matches the true area, so the work done by a variable force is equal to the area under the $F cos θ$ against $d$ curve.

Worked example: work from a force-displacement graph in three parts

The graph shows the force acting along the direction of motion of a trolley. Find the work done on the trolley as it moves the first

9 m

Force against displacement for a trolley

d / m

F / N

A_{1}

A_{2}

A_{3}

The force rises, stays steady, then falls to zero.

The work is the area under the graph, so split the area into a triangle, a rectangle and a second triangle.
First triangle (base $3 m$ , height $12 N$ ): area $= \frac{1}{2} \times base \times height$ .
$A_{1} = \frac{1}{2} \times 3 \times 12 = 18 J$
Rectangle (length $7 - 3 = 4 m$ , height $12 N$ ).
$A_{2} = 4 \times 12 = 48 J$
Second triangle (base $9 - 7 = 2 m$ , height $12 N$ ).
$A_{3} = \frac{1}{2} \times 2 \times 12 = 12 J$
Add the three areas.
$W = 18 + 48 + 12 = 78 J$
Check: the whole shape is a trapezoid with parallel sides $9 m$ and $4 m$ and height $12 N$ .
$\frac{1}{2} (9 + 4) \times 12 = 78 J$

W = 78 J

5.2Conservative and non-conservative fields

The region around a body where it can influence other bodies by a force associated with it is called a field of force or simply a force field. Examples include the electric field, the viscous field and the gravitational field.

Force field: The region around a body where it can influence other bodies by a force associated with it.

Conservative field

A field is said to be conservative if it has two important properties.

Work done is independent of the path taken. If the work done on a particle moving between any two points A and B is the same for path I and path II, as shown in Figure 5.6 (a), the field is conservative.
Work done around a closed path is zero. If the work done on a particle moving through any closed path (a path in which the beginning and end points are the same) is zero, as shown in Figure 5.6 (b), such a field is conservative.

Figure 5.6: (a) an object moves between points A and B via two different paths I and II. (b) The object makes a round trip back to A.

Gravitational and electric fields are examples of conservative fields, and the associated forces are conservative forces.

Non-conservative field

Non-conservative field: A field in which the work done depends upon the path followed, or the work done along a closed path is not zero.

A frictional field is a non-conservative field, and frictional or drag forces are non-conservative forces, because when an object is moved in a frictional field the work done against the frictional force depends upon the path followed. Frictional force, viscous drag and air resistance are all examples of non-conservative forces, and the fields where they act are termed non-conservative fields.

Conservative and non-conservative fields.
	Conservative field	Non-conservative field
Work between two points	independent of the path	depends on the path followed
Work around a closed path	zero	not zero
Examples	gravitational field, electric field	frictional field, viscous drag, air resistance

Worked example: gravity and friction on different routes

A 2.0\,\mathrm{kg} book is raised 1.5\,\mathrm{m} from the floor to a shelf, once straight up and once by a longer winding route. Separately, it is slid across a rough table where friction is 20\,\mathrm{N}, from A to B by the two paths of the figure. Find the work done by (a) gravity on each lift, (b) friction on each path, and (c) by each force over a round trip.

(a) Gravity acts down and the book rises 1.5\,\mathrm{m}, so only the vertical rise counts, whatever the route.
W_g=-mgh=-2.0\times9.8\times1.5=-29.4\,\mathrm{J}\ \text{(both routes)}
(b) Friction always opposes the motion, so \cos\theta=-1 and the work depends on the length of the path.
W_1=-20\times4.0=-80\,\mathrm{J},\qquad W_2=-20\times6.0=-120\,\mathrm{J}
(c) Gravity: the work down on the way back cancels the work up, so the closed-path total is zero. Friction: going A to B by path 1 and back by path 2 gives
W=-80-120=-200\,\mathrm{J}\neq0
Gravity gives the same work on every route and zero round a closed loop, so it is conservative. Friction gives different work on different routes and a non-zero total round a loop, so it is non-conservative.

Figure: two routes from A to B on a rough table. The longer route (accent) gives more work against friction.

W_g=-29.4\,\mathrm{J}\ \text{on both routes};\ W_f=-80\,\mathrm{J}\ \text{and}\ -120\,\mathrm{J}

5.3Kinetic energyOften examined

The energy possessed by a body due to its motion is called kinetic energy.

Kinetic energy: The energy possessed by a body due to its motion.

A boy kicks a football and it moves because it possesses kinetic energy. Think of a tennis ball and a football moving with the same speed: the football, with the larger mass, is harder to stop and can do more work. Now think of two footballs approaching you with different speeds: the one with the greater speed is harder to stop. Thus the object's mass and its speed both contribute to its kinetic energy. Like all energies, kinetic energy is a scalar quantity.

Derivation Kinetic energy from the equations of motion

Aim. Derive the formula for the kinetic energy of a moving body, using the equations of motion.

Assumptions.

All the work done transfers only kinetic energy to the body (a cricket ball).
The ball is initially at rest, v_i=0, and a constant horizontal force F moves it through a displacement d to a final velocity v_f=v (Figure 5.7).

Figure 5.7: work done transfers kinetic energy to a cricket ball.

The work done W appears as the kinetic energy K.E. Work is force times displacement.
W=\text{K.E.}=Fd\qquad(5.4)
Newton's second law gives the force.
F=ma\qquad(5.5)
The third equation of motion, v_f^{2}=v_i^{2}+2ad, is solved for the distance d.
d=\dfrac{v_f^{2}-v_i^{2}}{2a}\qquad(5.6)
Put Eqs. (5.5) and (5.6) in Eq. (5.4). The acceleration a cancels.
\text{K.E.}=ma\times\dfrac{v_f^{2}-v_i^{2}}{2a}=\dfrac12m\left(v_f^{2}-v_i^{2}\right)\qquad(5.7)
The ball started from rest, so v_i=0 and v_f=v.
\text{K.E.}=\dfrac12mv^{2}\qquad(5.8)

Result\text{K.E.}=\dfrac12mv^{2}

Although derived for a cricket ball, Eq. (5.8) gives the kinetic energy of any moving object. Kinetic energy is proportional to the mass and to the square of the speed. For example, an iron ball of 1.0\,\mathrm{kg} moving with a speed of 2.0\,\mathrm{m\,s^{-1}} has kinetic energy \tfrac12\times1.0\times2.0^{2}=2\,\mathrm{J}.

Equally important, the derivation demonstrates the work-kinetic energy theorem, which states that the work done on an object is equal to the change in its energy, i.e. W=\Delta E, where W is the work done and \Delta E is the change in energy.

W=\Delta E

5.4Work-energy principleOften examined

Since work is defined as the movement of an object through a distance, energy can also be described as the ability to move an object through a distance. The net work done on an object is equal to the change in the object's kinetic energy.

Work-energy principle (theorem): The net work done on an object is equal to the change in its kinetic energy: W_{net}=\text{K.E.}_f-\text{K.E.}_i=\Delta\text{K.E.}

W_{net}=\text{K.E.}_f-\text{K.E.}_i=\Delta\text{K.E.}

5.9

Here the change in the kinetic energy is due to the object's change in speed. Using \text{K.E.}=\tfrac12mv^{2}, the equation reads W_{net}=\tfrac12mv_f^{2}-\tfrac12mv_i^{2}.

The work-energy theorem is also valid for potential energy. However, potential energy can be defined only for conservative forces. In situations where potential energy can be defined, the change in potential energy is exactly equal to the negative of the change in kinetic energy, in which case the work-energy theorem becomes

W_{net}=\Delta\text{K.E.}=-\Delta U

5.10

Worked example: final speed from the work-energy principle

A 4.0\,\mathrm{kg} block moving at 3.0\,\mathrm{m\,s^{-1}} on a smooth surface is pushed by a net force of 12\,\mathrm{N} along its motion for 5.0\,\mathrm{m}. Find its final speed and check with the equations of motion.

Given

m = 4.0\,\mathrm{kg}
v_i = 3.0\,\mathrm{m\,s^{-1}}
F_{net} = 12\,\mathrm{N}
d = 5.0\,\mathrm{m}

To find

final speed v_f

The net work is the net force times the displacement (same direction).
W_{net}=F_{net}d=12\times5.0=60\,\mathrm{J}
Initial kinetic energy.
\text{K.E.}_i=\tfrac12(4.0)(3.0)^2=18\,\mathrm{J}
By the work-energy principle W_{net}=\text{K.E.}_f-\text{K.E.}_i.
\text{K.E.}_f=18+60=78\,\mathrm{J}
Solve \tfrac12mv_f^2=78 for the speed.
v_f=\sqrt{\dfrac{2\times78}{4.0}}=\sqrt{39}=6.2\,\mathrm{m\,s^{-1}}
Check: a=F/m=3.0\,\mathrm{m\,s^{-2}} and v_f^2=v_i^2+2ad.
v_f^2=9+2(3.0)(5.0)=39\ \Rightarrow\ v_f=6.2\,\mathrm{m\,s^{-1}}

v_f\approx6.2\,\mathrm{m\,s^{-1}}

5.4.1Work-energy theorem in resistive medium

Why is a skydiver quite confident when using a parachute? Probably she relies on a balance between resistive forces and her weight. A resistive force on a moving object opposes the motion of the object or prevents a stationary object from moving, for example friction or viscous force.

Resistive force: A force on a moving object that opposes the motion of the object, or prevents a stationary object from moving, for example friction or viscous force.

When forces act on an object, energy transformations occur, which means work is being done. The work done W on an object by an applied force F is the sum of the gain in kinetic energy \Delta\text{K.E.} of the object and the work W_r done by the object against the resistive force. That is:

W=\Delta\text{K.E.}+W_r

The work done by the object against the resistive force takes energy away from the object, decreasing its kinetic energy. Rearranging, mathematically:

W_r=W-\Delta\text{K.E.}

5.11

Both Newton's laws and the work-energy theorem can be used to solve problems. However, if the forces are not constant, Newton's laws will be difficult to apply, while the work-energy theorem still works because it only needs the work done and the change in kinetic energy.

Worked example: where the pusher's work goes

A boy pushes a 30\,\mathrm{kg} cart from rest with a steady 60\,\mathrm{N} force over 10\,\mathrm{m}. The cart ends at 4.0\,\mathrm{m\,s^{-1}}. Find the work done against resistance and the average resistive force.

Work done by the boy.
W=Fd=60\times10=600\,\mathrm{J}
Change in kinetic energy (started from rest).
\Delta\text{K.E.}=\tfrac12(30)(4.0)^2-0=240\,\mathrm{J}
From W=\Delta\text{K.E.}+W_r.
W_r=W-\Delta\text{K.E.}=600-240=360\,\mathrm{J}
The resistive force acts over the same 10\,\mathrm{m}.
F_r=\dfrac{W_r}{d}=\dfrac{360}{10}=36\,\mathrm{N}
Check with Newton's law: net force =60-36=24\,\mathrm{N}, a=0.80\,\mathrm{m\,s^{-2}}, so v^2=2ad.
v^2=2(0.80)(10)=16\ \Rightarrow\ v=4.0\,\mathrm{m\,s^{-1}}

Figure: the 600 J of work splits into kinetic energy gained and work spent against the resistive force (accent).

W_r=360\,\mathrm{J},\quad F_r=36\,\mathrm{N}

5.4.2Implications of energy loses in practical devices and efficiencyOften examined

Input, output and efficiency of a machine

Input of a mechanical machine: The energy supplied to a mechanical machine. It is the product of the effort P and the distance d through which the effort acts.
Output of a mechanical machine: The work done by a mechanical machine. It is the product of the load W and the distance D through which the load lifts.
Efficiency: The ratio of the work output to the work input (useful energy output divided by energy input).

\text{input}=P\times d

\text{output}=W\times D

\text{efficiency}=\dfrac{\text{useful energy output}}{\text{energy input}}

Figure 5.8: input and output of a pulley. The effort P moves through d (input =P\times d); the load W rises through D (output =W\times D).

Efficiency is the ratio of similar quantities and therefore has no unit. The efficiency cannot be greater than 1 (or 100\%); in fact it cannot be equal to 1 (or 100\%) for real machines. In the case of a pulley system the efficiency is only 40\% and the rest, 60\%, goes into waste and is converted into unwanted forms of energy.

Ideal machine

For an ideal machine only, the input will be equal to the output. This means that no energy is wasted and all energy is converted into useful work. Therefore the efficiency of an ideal machine is 100\%.

\text{output}=\text{input}\qquad\text{or}\qquad\text{efficiency}=1

Energy losses in practical devices

Almost all energy transformation technologies operate at efficiencies less than 100\%, and most of the wasted energy becomes thermal energy. For example, automobiles are highly inefficient. Suppose an amount of fuel containing 1000\,\mathrm{J} of chemical potential energy is used by an automobile's engine and only 10\% of the energy is available to do work.

\text{Efficiency}=\dfrac{\text{energy output}}{\text{energy input}}\times100=\dfrac{100\,\mathrm{J}}{1000\,\mathrm{J}}\times100=10\%

Only about 10\% (or 100\,\mathrm{J}) of useful work is done in producing the kinetic energy of the moving car. The rest of the energy is wasted in the engine and the car's transmission. Some other examples of energy losses in practical devices are given in Table 5.1.

Table 5.1: typical efficiencies of energy transformation technologies.
Device	Efficiency (%)
Electric generator	98
Hydroelectric power plant	95
Large electric motor	95
Home gas furnace	85
Wind generator	55
Fossil fuel power plant	40
Automobile engine	25
Fluorescent light	20
Incandescent light	5

Example 5.2

In a science experiment, a 3.00\,\mathrm{kg} water rocket is launched from the ground. The rocket's total energy at the top of its flight is 2352\,\mathrm{J}. (a) What was the rocket's launching speed? (b) What height did the rocket reach? (c) What are the kinetic energy and potential energy of the rocket 2.5\,\mathrm{s} after its launch?

Given

m = 3.00\,\mathrm{kg} mass of rocket
E_T = 2352\,\mathrm{J} total energy
g = 9.8\,\mathrm{m\,s^{-2}}
t = 2.5\,\mathrm{s}

To find

(a) initial speed v
(b) height h
(c) K.E. and P.E. after 2.5\,\mathrm{s}

(a) At launch all the energy is kinetic. By the law of conservation of energy this kinetic energy is converted into potential energy at the top, which equals the total energy.
\text{K.E.}=E_T\quad\Rightarrow\quad\tfrac12mv^{2}=E_T\quad\Rightarrow\quad v^{2}=\dfrac{2E_T}{m}
Take the square root and put in the values.
v=\sqrt{\dfrac{2E_T}{m}}=\sqrt{\dfrac{2\times2352}{3.00}}=39.6\,\mathrm{m\,s^{-1}}
(b) At the greatest height the potential energy equals the total energy.
\text{P.E.}=E_T\quad\Rightarrow\quad mgh=E_T\quad\Rightarrow\quad h=\dfrac{E_T}{mg}
Put in the values.
h=\dfrac{2352}{3.00\times9.80}=80\,\mathrm{m}
(c) First find the speed after 2.5\,\mathrm{s} with the first equation of motion along the vertical axis. The rocket moves up, so gravity decreases the speed.
v_f=v_i-gt=39.6-9.8\times2.5=15.1\,\mathrm{m\,s^{-1}}
Put this speed in the kinetic energy equation.
\text{K.E.}=\tfrac12mv_f^{2}=\tfrac12\times3.00\times(15.1)^{2}=342\,\mathrm{J}
For the potential energy use conservation of energy, E_T=\text{K.E.}+\text{P.E.}
\text{P.E.}=E_T-\text{K.E.}=2352-342=2010\,\mathrm{J}

v=39.6\,\mathrm{m\,s^{-1}},\quad h=80\,\mathrm{m},\quad \text{K.E.}=342\,\mathrm{J},\quad \text{P.E.}=2010\,\mathrm{J}

Assignment 5.2

A crane holding 5000\,\mathrm{kg} of mass at a height of 12\,\mathrm{m} suddenly un-holds the mass. Find the kinetic energy of the mass just before striking the ground.

Given

m = 5000\,\mathrm{kg}
h = 12\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

K.E. just before striking the ground

At 12\,\mathrm{m} the mass is at rest, so all its energy is potential.
\text{P.E.}=mgh=5000\times9.8\times12=5.88\times10^{5}\,\mathrm{J}
Neglecting air resistance, by conservation of energy (or the work-energy theorem, W=mgh=\Delta\text{K.E.}) all this potential energy becomes kinetic energy just before the mass strikes the ground, where the P.E. is zero.
\text{K.E.}=\text{P.E.}=5.88\times10^{5}\,\mathrm{J}

\text{K.E.}=5.88\times10^{5}\,\mathrm{J}\approx588\,\mathrm{kJ}

Formula sheet

Work done by a constant force

W=\vec F\cdot\vec d=Fd\cos\theta

F: magnitude of force; d: magnitude of displacement; \theta: angle between \vec F and \vec d (Eq. 5.1)

Unit: \mathrm{J}

The joule

1\,\mathrm{J}=1\,\mathrm{N\,m}

work of 1\,\mathrm{N} through 1\,\mathrm{m} along the force

Unit: \mathrm{J}

Total work by a variable force

W_{Total}=\sum_{i=1}^{n}F_i\cos\theta_i\,\Delta d_i

F_i\cos\theta_i: effective force over the small displacement \Delta d_i; equals the area under the F\cos\theta against d graph (Eq. 5.3)

Unit: \mathrm{J}

Kinetic energy

\text{K.E.}=\dfrac12mv^{2}

m: mass; v: speed (Eq. 5.8)

Unit: \mathrm{J}

Work-kinetic energy theorem

W=\Delta E

W: work done; \Delta E: change in energy

Unit: \mathrm{J}

Work-energy principle

W_{net}=\text{K.E.}_f-\text{K.E.}_i=\Delta\text{K.E.}

W_{net}: net work done on the object (Eq. 5.9)

Unit: \mathrm{J}

Work-energy theorem with potential energy

W_{net}=\Delta\text{K.E.}=-\Delta U

\Delta U: change in potential energy, for conservative forces only (Eq. 5.10)

Unit: \mathrm{J}

Work against a resistive force

W_r=W-\Delta\text{K.E.}

W: work done by the applied force; \Delta\text{K.E.}: gain in kinetic energy; W_r: work done against the resistive force (Eq. 5.11)

Unit: \mathrm{J}

Input of a machine

\text{input}=P\times d

P: effort; d: distance through which the effort acts

Unit: \mathrm{J}

Output of a machine

\text{output}=W\times D

W: load; D: distance through which the load lifts

Unit: \mathrm{J}

Efficiency

\text{efficiency}=\dfrac{\text{useful energy output}}{\text{energy input}}\ \ (\times100\text{ for a percentage})

less than 1 (100\%) for a real machine; 1 for an ideal machine

no unit

Key terms

Work: The scalar (dot) product of force and displacement, W=Fd\cos\theta.
Joule: The SI unit of work; 1\,\mathrm{J}=1\,\mathrm{N\,m}.
Effective component: The component of the force along the displacement, F\cos\theta; the perpendicular component F\sin\theta does no work.
Positive work: Work done when the force has a component along the displacement (0^\circ\le\theta<90^\circ).
Negative work: Work done when the force has a component opposite to the displacement (90^\circ<\theta\le180^\circ).
Zero work: Work done by a force perpendicular to the displacement (\theta=90^\circ).
Force field: The region around a body where it can influence other bodies by a force associated with it.
Conservative field: A field in which the work done is independent of the path taken and the work done around a closed path is zero, for example the gravitational and electric fields.
Non-conservative field: A field in which the work done depends upon the path followed or the work done along a closed path is not zero, for example a frictional field.
Kinetic energy: The energy possessed by a body due to its motion, \tfrac12mv^{2}.
Work-kinetic energy theorem: The work done on an object is equal to the change in its energy, W=\Delta E.
Work-energy principle: The net work done on an object is equal to the change in its kinetic energy, W_{net}=\Delta\text{K.E.}
Resistive force: A force that opposes the motion of a moving object or prevents a stationary object from moving, for example friction or viscous force.
Input (of a machine): The energy supplied to a mechanical machine: effort times the distance through which it acts.
Output (of a machine): The work done by a mechanical machine: load times the distance through which the load lifts.
Efficiency: The ratio of the work output to the work input; it has no unit and is 100\% only for an ideal machine.
Ideal machine: A machine in which no energy is wasted, so output equals input and the efficiency is 100\%.

Multiple choice questionsp. 85–86

If the unit of force and displacement travelled each be increased five times, then the unit of work will be increased by:

A25 timesAnswer
B10 times
C5 times
D0 times

Why

Work is the product of force and displacement, W=Fd\cos\theta. If F and d are each multiplied by 5, the new work is (5F)(5d)=25\,Fd, which is 25 times the original work. Option A is correct.

Option C (5 times) counts only one of the two changes. Work depends on the product of force and displacement, so both changes multiply.

Note. The printed wording 'unit of force and displacement travelled each be increased' is loose; it is read as the magnitudes of the force and the displacement being increased five times. The book prints no key.

The force that acts on a body but does no work is:

Agravitational force
Bfrictional force
Celastic force
Dcentripetal forceAnswer

Why

A centripetal force points towards the centre of the circle, so it is perpendicular to the velocity and to the small displacement along the path. The angle is \theta=90^\circ and W=Fd\cos90^\circ=0.

Gravitational, frictional and elastic forces can each do work when the displacement has a component along the force, for example friction on a sliding block does negative work.

The odd force from the following is:

Agravitational force
Belastic force
Cfrictional forceAnswer
Delectric force

Why

Gravitational, elastic and electric forces are conservative: the work they do does not depend on the path, and the work around a closed path is zero, so a potential energy can be defined for each.

Frictional force is non-conservative, because the work done against friction depends on the path followed (Section 5.2.2). It is the odd one out.

The work done by a body while covering a vertical height of 10 m is 500 J. By how much amount does the energy of the body change?

A500\,\mathrm{J}
B-500\,\mathrm{J}Answer
C50\,\mathrm{J}
D-50\,\mathrm{J}

Why

Work done by the body is paid for from its own energy, so the energy of the body falls by the same amount: \Delta E=-W=-500\,\mathrm{J}. The energy change is negative because the energy decreases.

Option A gives the size of the work but not the sign. Options C and D use the wrong number. The 10 m height is not needed to find the energy change.

52025 Annual I, Set B, MCQ 62025 Annual I, Set D, MCQ 5

Two objects, A and B, have the same mass. Object A is moving at twice the speed of object B. The kinetic energy of object A as compared to object B is

AK.E_A=4\,K.E_BAnswer
BK.E_A=2\,K.E_B
CK.E_A=K.E_B
DK.E_A=\tfrac12\,K.E_B

Why

\frac{K.E_A}{K.E_B}=\frac{\tfrac12 m(2v)^2}{\tfrac12 mv^2}=\frac{2mv^2}{\tfrac12 mv^2}=4

Kinetic energy depends on the square of the speed, so doubling the speed multiplies it by 2^2=4. Option B forgets the square.

If velocity of a body reduces to half of its initial value, then the kinetic energy of the body compared to its initial value is:

Aone fourthAnswer
Bdouble
Cfour times
Dhalf

Why

K.E'=\tfrac12 m\left(\tfrac{v}{2}\right)^2=\tfrac14\left(\tfrac12 mv^2\right)=\tfrac14\,K.E

Kinetic energy is proportional to v^2, so halving the speed leaves one quarter of the energy. Option D (half) treats kinetic energy as proportional to speed, which is wrong.

A moving hockey ball is hit, such that its speed doubles. Its kinetic energy will become

Adouble
BquadrupleAnswer
Chalve
Dzero

Why

K.E'=\tfrac12 m(2v)^2=4\left(\tfrac12 mv^2\right)=4\,K.E

The kinetic energy becomes four times as large. Doubling the speed would double the energy only if kinetic energy were proportional to speed, and it is not.

An engine converts 600 J of input energy into 150 J of useful work. The efficiency of the engine is

A25 %Answer
B50 %
C75 %
D100 %

Why

\eta=\frac{\text{useful energy output}}{\text{energy input}}\times100\,\%=\frac{150\,\mathrm{J}}{600\,\mathrm{J}}\times100\,\%=25\,\%

Option C (75 %) is the fraction wasted, 450\,\mathrm{J} out of 600\,\mathrm{J}, not the efficiency.

Two engines, A and B, are rated for the same power output. Engine A has a higher efficiency than engine B. What can you conclude?

AEngine A requires less fuel to produce the same power.Answer
BEngine A is physically larger and heavier than engine B.
CEngine B produces more waste heat than engine A.
DThe efficiency rating is unrelated to fuel consumption.

Why

For the same useful output, \eta=P_{\text{out}}/P_{\text{in}} is larger when the input power is smaller. A higher efficiency therefore means less fuel energy is needed to deliver the same power. Option A is the conclusion.

B does not follow, since efficiency says nothing about size or weight. D is false, because efficiency links the energy taken in from the fuel to the useful energy given out.

Option C is also consistent with the data: for the same useful output, the less efficient engine must waste more energy as heat. A is the direct consequence about fuel that the question asks for, so it is the key.

Note. Option C is also true for these two engines, so the question has two consistent conclusions. The key is A because it is the direct consequence about fuel. The book prints no key.

Consider two machines, A and B. Machine A has an efficiency of 40 % and Machine B has an efficiency of 60 %. If both machines are provided with 500 J of energy, how much more useful work does Machine B perform compared to Machine A?

A50 J
B100 JAnswer
C200 J
D300 J

Why

W_A=0.40\times500\,\mathrm{J}=200\,\mathrm{J},\qquad W_B=0.60\times500\,\mathrm{J}=300\,\mathrm{J}

The difference is 300\,\mathrm{J}-200\,\mathrm{J}=100\,\mathrm{J}, so B is correct. Option C is the useful work of A alone and option D is the useful work of B alone; neither is the difference.

A wind turbine converts 30 % of the wind's kinetic energy into electrical energy. If the kinetic energy of the wind is 900 J. The electrical energy produced by the turbine is

A90 J
B270 JAnswer
C600 J
D700 J

Why

E_{\text{el}}=0.30\times900\,\mathrm{J}=270\,\mathrm{J}

Option A (90 J) would correspond to a conversion of only 10 %, which is not what the question states.

Short questionsp. 86

5.1

Is it possible that a force is acting on a body and the body is in motion due to this force but the work done after certain time is zero?

Solution

Yes. Work is W=Fd\cos\theta, so the work is zero whenever the force is perpendicular to the displacement (\theta=90^\circ), because \cos90^\circ=0.

A stone whirled in a horizontal circle at constant speed is an example. The string tension acts towards the centre, perpendicular to the displacement at every instant, so the work done is zero although the stone is moving. The force changes only the direction of motion, not the speed.

A second case is a conservative force acting over a closed path. If the body returns to its starting point, the total work is zero, even though the body moved and the force acted.

5.2

Some non-conservative forces are acting on a body. Can they change the total mechanical energy of the body?

Solution

Yes. The total mechanical energy is the sum of kinetic and potential energy. Non-conservative forces such as friction, viscous drag and air resistance do work that depends on the path, and this work changes the mechanical energy.

W_{\text{nc}}=W_{\text{net}}-W_{\text{c}}=\Delta K.E-(-\Delta U)=\Delta K.E+\Delta U=\Delta E_{\text{mech}}

Here W_{\text{c}}=-\Delta U is the work done by the conservative forces, and W_{\text{nc}} is the work done by the non-conservative forces. So the work done by non-conservative forces equals the change in the total mechanical energy.

For example, a block sliding on a rough table loses kinetic energy to friction. Its total mechanical energy falls, and the energy is converted into other, mainly thermal, forms. The book's form of this idea is W_r=W-\Delta K.E (5.11).

5.3

Kinetic energy and work are related. Can kinetic energy ever be negative? Can work ever be negative?

Solution

Kinetic energy cannot be negative. K.E=\tfrac12 mv^2 with a positive mass m and v^2\ge0, so K.E\ge0. It is zero only when the body is at rest.

Work can be negative. W=Fd\cos\theta is negative when 90^\circ<\theta\le180^\circ, that is, when the force has a component opposite to the displacement. Friction on a sliding block and the braking force of a car do negative work. Negative work means energy is taken away from the body.

5.42026 Annual I, Set B, Q.2 (iv)2025 Annual I, Set B, Q.2 (iii)2025 Annual I, Set D, Q.2 (v) OR

Differentiate between conservative and non-conservative forces.

Solution

A conservative force is one whose work depends only on the initial and final positions, not on the path. A non-conservative force does work that depends on the path followed.

Conservative and non-conservative forces
Property	Conservative force	Non-conservative force
Work along a path	Independent of the path	Depends on the path followed
Work around a closed path	Zero	Not zero
Potential energy	Can be defined	Cannot be defined
Mechanical energy	Conserved	Not conserved
Examples	Gravitational, electric	Friction, viscous drag, air resistance

The fields in which they act are called conservative fields (gravitational and electric fields) and non-conservative fields (frictional and viscous fields).

5.5

What is the work done by the moon as it revolves around the Earth?

Solution

The work is zero. The moon's orbit is treated as circular. The only force on the moon is the gravitational pull of the Earth, which always points towards the Earth's centre and so is perpendicular to the moon's velocity at every instant. Then \theta=90^\circ and W=Fd\cos90^\circ=0.

The moon's speed stays constant, so its kinetic energy does not change, which agrees with the work-energy theorem. Over one full revolution the work is also zero, because gravity is conservative and the path is closed, so the result holds even if the orbit is slightly elliptical.

W=0

Note. The question asks for the work done by the moon, which is the work done on the moon by the Earth's gravitational pull. The book gives no answer. The orbit is taken as circular.

Comprehensive questionsp. 86–87

5.12026 Annual I, Set B, Q.62025 Annual I, Set D, Q.2 (vi) OR

Derive the formula for kinetic energy by using the equations of motion.

Solution

Kinetic energy is the energy a body possesses because of its motion. Take a ball, initially at rest, which is given a horizontal force F so that it moves through a displacement d and reaches a final speed v. The work done on the ball appears as its kinetic energy (Figure 5.7).

Derivation Kinetic energy from the equations of motion

Aim. Derive the kinetic energy of a body of mass m that starts from rest and reaches speed v under a constant force.

Assumptions.

The force F is constant and acts in the direction of the displacement.
The mass m is constant and no other force, such as friction, does work on the ball.

Figure 5.7: Work done transfers kinetic energy to a cricket ball.

Write the work done by the force over the displacement. The force and displacement are in the same direction, so \theta=0^\circ.
W=Fd\cos0^\circ=Fd
Newton's second law gives the force in terms of the acceleration a it produces.
F=ma
Substitute the force into the work, so that the work depends on m, a and d.
W=mad
Use the third equation of motion, which links the initial speed v_i, the final speed v_f, the acceleration and the displacement. Solve it for d.
v_f^2=v_i^2+2ad\quad\Rightarrow\quad d=\dfrac{v_f^2-v_i^2}{2a}
Substitute this d into W=mad. The acceleration cancels.
W=ma\times\dfrac{v_f^2-v_i^2}{2a}=\dfrac12 m\left(v_f^2-v_i^2\right)
The work is equal to the energy gained by the ball. Write K.E=\tfrac12 mv^2 for a body of speed v, so the gain is K.E_f-K.E_i.
W=K.E_f-K.E_i
The ball starts from rest, so v_i=0 and K.E_i=0. Write v_f=v.
W=\dfrac12 mv^2

ResultK.E=\tfrac12 mv^2

Kinetic energy depends on the mass and on the square of the speed. An iron ball of mass 1.0\,\mathrm{kg} moving with speed 2.0\,\mathrm{m\,s^{-1}} has K.E=\tfrac12\times1.0\times2.0^2=2\,\mathrm{J}, as the book states.

Since \mathrm{kg\,m^{2}\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}, the SI unit of kinetic energy is the joule. Although derived for a cricket ball, the formula holds for any moving body.

5.22025 Annual I, Set D, Q.2 (iv) OR

Explain work done by a constant and variable force using force-displacement graph.

Solution

Work done by a constant force

A constant force acting along the displacement has the same value at every point, so its force-displacement graph is a horizontal line. The work done is the area of the rectangle between the line and the displacement axis: W=F\times d (Figure 5.3).

Constant force

Figure 5.3: Work done by a constant force is the area of the rectangle under the force-displacement graph.

Example: a constant force of 5\,\mathrm{N} acts along a displacement of 6\,\mathrm{m}. The area is the work done.
W=F\,d=5\,\mathrm{N}\times6\,\mathrm{m}=30\,\mathrm{J}

Work done by a variable force

When the force changes with displacement, the displacement is divided into small intervals \Delta d_1,\Delta d_2,\ldots over which the force is almost constant. In each interval the work is F_i\cos\theta_i\,\Delta d_i, which is the area of one thin rectangle. The total work is the sum of these areas (5.2), and it approaches the exact area under the F\cos\theta and d graph as the intervals become thinner.

W_{\text{Total}}=\sum_{i=1}^{n}F_i\cos\theta_i\,\Delta d_i

When the force increases uniformly with displacement, the graph is a straight line and the area under it is a triangle (Figure 5.4).

Variable force

Figure 5.4: Work done by a variable force is the area of the region under the force-displacement graph.

Example: the force rises uniformly from 0 to 6\,\mathrm{N} over 5\,\mathrm{m}. The area under the graph is a triangle, whose area is half the base times the height.
W=\tfrac12\times d\times F_{\max}
Substitute d=5\,\mathrm{m} and F_{\max}=6\,\mathrm{N}.
W=\tfrac12\times5\,\mathrm{m}\times6\,\mathrm{N}=15\,\mathrm{J}

In every case the work done is the area between the force-displacement graph and the displacement axis. The method works for any shape of graph, as the area is the sum of the strips.

5.32026 Annual I, Set B, Q.62025 Annual I, Set B, Q.4 OR

What is work-energy theorem? Explain in detail. Also, write some implications of energy loses in practical devices and efficiency.

Solution

Work-energy theorem

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_{\text{net}}=K.E_f-K.E_i=\Delta K.E. The change in kinetic energy is caused by the change in the object's speed.

Derivation Work-energy theorem

Aim. Show that the net work done on a body equals its change in kinetic energy.

Assumptions.

The net force F_{\text{net}} is constant and acts along the displacement d.
The mass m is constant.

The net force does work over the displacement d in its own direction.
W_{\text{net}}=F_{\text{net}}\,d
Newton's second law gives the net force.
F_{\text{net}}=ma
Use the third equation of motion and solve it for d.
d=\dfrac{v_f^2-v_i^2}{2a}
Substitute the net force and d, and cancel a.
W_{\text{net}}=ma\,\dfrac{v_f^2-v_i^2}{2a}=\dfrac12 mv_f^2-\dfrac12 mv_i^2
Each term is a kinetic energy, so the right side is the change in kinetic energy.
W_{\text{net}}=K.E_f-K.E_i=\Delta K.E

ResultW_{\text{net}}=\Delta K.E

Positive net work increases the kinetic energy, so the body speeds up. Negative net work decreases it, so the body slows down. The theorem gives the answer without finding the acceleration, which is useful when Newton's laws are difficult to apply because the force is not constant.

Work-energy theorem with conservative forces

When the forces are conservative, a potential energy can be defined. The change in potential energy is the negative of the work done by these forces, so the theorem becomes W_{\text{net}}=\Delta K.E=-\Delta U (5.10). The sum of kinetic and potential energy then stays constant when only conservative forces act.

Work-energy theorem in a resistive medium

A resistive force, such as friction or viscous force, opposes the motion. Part of the work done by the applied force W goes into the gain in kinetic energy, and the rest is the work W_r done by the object against the resistive force, so W=\Delta K.E+W_r (5.11).

The work done by the object against the resistive force takes energy away from the object and decreases its kinetic energy. This is why a parachute, which increases the resistive force, makes a skydiver slow down.

Energy losses and efficiency

The energy supplied to a machine is its input, equal to the product of the effort p and the distance d through which it acts: \text{input}=p\times d. The work done by the machine is its output, equal to the load W times the distance D through which the load lifts: \text{output}=W\times D.

\text{efficiency}=\frac{\text{useful energy output}}{\text{energy input}}\times100\,\%

Efficiency is the ratio of two similar quantities, so it has no unit. It cannot be greater than 1 (or 100\,\%), and for a real machine it is always less. In the pulley of Figure 5.8 the efficiency is only 40\,\%, and the other 60\,\% is wasted as unwanted forms of energy.
For an ideal machine, all the input becomes useful output, so the efficiency is 100\\,\\% (output = input).
\text{output}=\text{input}\quad\Rightarrow\quad\text{efficiency}=1
Example: an automobile engine uses fuel with 1000\\,\\mathrm{J} of chemical energy and only 100\\,\\mathrm{J} is useful work.
\text{efficiency}=\frac{100\,\mathrm{J}}{1000\,\mathrm{J}}\times100\,\%=10\,\%

Table 5.1: Typical efficiencies of energy transformation technologies
Device	Efficiency (%)
Electric generator	98
Hydroelectric power plant	95
Large electric motor	95
Home gas furnace	85
Wind generator	55
Fossil fuel power plant	40
Automobile engine	25
Fluorescent light	20
Incandescent light	5

Almost all energy transformation technologies operate at efficiencies below 100\,\%. The wasted energy usually becomes thermal energy, which cannot all be turned back into useful work. This is the practical implication of energy losses: each device delivers less useful work than the energy it receives.

Numerical problemsp. 87

5.1

A car carrying truck unloads a 1500 kg car using a plank, as shown in figure. If the plank makes an angle of 30° with the ground and its upper end is at 2 m height, what will be the work done by gravitational force? Also, draw the force-displacement graph for.

Solution

Given

m = 1500\,\mathrm{kg} car
\theta = 30^\circ angle of the plank with the ground
h = 2\,\mathrm{m} height of the upper end
g = 9.8\,\mathrm{m\,s^{-2}}

To find

W_g, the work done by gravity
the force-displacement graph along the plank

The car slides down a plank inclined at 30^\circ from a height of 2\,\mathrm{m} to the ground.

Find the length of the plank. The car moves along the plank, so the displacement d is the length of the plank.
\sin30^\circ=\frac{h}{d}\quad\Rightarrow\quad d=\frac{h}{\sin30^\circ}=\frac{2}{0.5}=4\,\mathrm{m}
Find the component of the weight along the plank. This is the force that does the work; the other component is balanced by the plank.
F=mg\sin30^\circ=1500\times9.8\times0.5=7350\,\mathrm{N}
The force is constant and acts along the displacement, so the work is force times displacement.
W_g=F\,d=7350\,\mathrm{N}\times4\,\mathrm{m}=29\,400\,\mathrm{J}
Check with the height. Gravity depends only on the vertical drop, so the same value comes from mgh.
mgh=1500\times9.8\times2=29\,400\,\mathrm{J}

Force along the plank

The constant force mg\sin30^\circ=7350\,\mathrm{N} acts over 4\,\mathrm{m}. The shaded area is the work done by gravity, 29.4\,\mathrm{kJ}.

W_g=mgh=29.4\,\mathrm{kJ}\quad(\text{positive, because the car moves down the plank})

Book answer. 29.4\,\mathrm{kJ}

Note. The book prints 29.4 kJ without a sign. The car moves down the plank, so gravity does positive work. The book's last sentence, 'draw the force-displacement graph for.', is incomplete; the graph is drawn for the component of the weight along the plank, mg\sin30^\circ, which is the force that does the work.

5.2

A car, having a total mass of 1500 kg (including the driver), is travelling at a speed of 40 kph through a straight path. How much work will be required to stop the said car if its brakes fail and engine turns off?

Solution

Given

m = 1500\,\mathrm{kg} total mass, including the driver
v = 40\,\mathrm{km\,h^{-1}} speed

To find

W, the work required to stop the car

Convert the speed to SI units.
v=\frac{40\times1000}{3600}=11.11\,\mathrm{m\,s^{-1}}
The car ends at rest, so the work-energy theorem gives the work as the change in kinetic energy, W=0-K.E_i. The stopping force must remove this energy.
K.E_i=\tfrac12 mv^2
Evaluate the kinetic energy.
K.E_i=\tfrac12\times1500\times(11.11)^2=9.26\times10^{4}\,\mathrm{J}

|W|=92.6\,\mathrm{kJ}

Book answer. 92.6\,\mathrm{kJ}

Note. The book gives the magnitude. The work done on the car by the stopping force is negative, W=-92.6\,\mathrm{kJ}, because it removes the kinetic energy.

5.3

A ball of mass 100 g is released from a height of 30 m. If the ball encounters an air resistance of 0.4 N, find the kinetic energy of the ball just before striking the ground.

Solution

Given

m = 100\,\mathrm{g}=0.100\,\mathrm{kg}
h = 30\,\mathrm{m}
F_r = 0.4\,\mathrm{N} air resistance
g = 9.8\,\mathrm{m\,s^{-2}}

To find

K.E just before striking the ground

The work done by gravity over the fall: the ball moves down 30\,\mathrm{m} in the direction of its weight.
W_g=mgh=0.100\times9.8\times30=29.4\,\mathrm{J}
The air resistance acts upwards, opposite to the motion, so \theta=180^\circ and the work is negative.
W_r=-F_r\,h=-0.4\times30=-12\,\mathrm{J}
The ball starts from rest, so the work-energy theorem gives its kinetic energy as the sum of the work done.
K.E=W_g+W_r=29.4-12=17.4\,\mathrm{J}

K.E=17.4\,\mathrm{J}

Book answer. 17.4\,\mathrm{J}

5.4

(a) What is the kinetic energy of a car with a mass of 1200 kg travelling at a speed of 20\,\mathrm{m\,s^{-1}}? (b) If the car comes to a stop over a distance of 50 meters due to braking, what is the average force exerted by the brakes?

Solution

Given

m = 1200\,\mathrm{kg} car
v = 20\,\mathrm{m\,s^{-1}} speed
d = 50\,\mathrm{m} stopping distance

To find

(a) K.E
(b) the average braking force F

(a) What is the kinetic energy of the car?

Use the kinetic energy formula.
K.E=\tfrac12 mv^2=\tfrac12\times1200\times(20)^2
Evaluate.
K.E=2.4\times10^{5}\,\mathrm{J}=240\,\mathrm{kJ}

K.E=240\,\mathrm{kJ}

(b) If the car comes to a stop over 50 m due to braking, what is the average force exerted by the brakes?

The brakes do negative work W=-F\\,d and the car loses all its kinetic energy, so the work equals -K.E.
-F\,d=-K.E\quad\Rightarrow\quad F=\frac{K.E}{d}
Substitute the kinetic energy from part (a) and d=50\,\mathrm{m}.
F=\frac{2.4\times10^{5}\,\mathrm{J}}{50\,\mathrm{m}}=4800\,\mathrm{N}

F=4800\,\mathrm{N}

K.E=240\,\mathrm{kJ},\qquad F=4800\,\mathrm{N}

Book answer. 240\,\mathrm{kJ}, 4800\,\mathrm{N}

5.5

A roller coaster cart weighing 500 kg is at the peak of a 30 m tall hill and is travelling at 5\,\mathrm{m\,s^{-1}}. Find the total mechanical energy at the top of the hill. If it descends to the bottom without any friction losses, what will be its speed at the bottom?

Solution

Given

m = 500\,\mathrm{kg} mass of the cart
h = 30\,\mathrm{m} height of the hill
v = 5\,\mathrm{m\,s^{-1}} speed at the top
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) the total mechanical energy at the top
(b) the speed at the bottom

(a) Find the total mechanical energy at the top of the hill.

Kinetic energy at the top.
K.E=\tfrac12 mv^2=\tfrac12\times500\times5^2=6250\,\mathrm{J}
Potential energy at the top, taking the bottom as the zero level.
P.E=mgh=500\times9.8\times30=1.47\times10^{5}\,\mathrm{J}
Total mechanical energy is the sum.
E=K.E+P.E=6250+147\,000=1.5325\times10^{5}\,\mathrm{J}

E=1.5325\times10^{5}\,\mathrm{J}

(b) If it descends to the bottom without friction losses, what will be its speed at the bottom?

There is no friction, so the total mechanical energy is conserved. At the bottom the potential energy is zero, so all the energy is kinetic.
\tfrac12 mv_b^2=E
Solve for the speed at the bottom.
v_b^2=\frac{2E}{m}=\frac{2\times1.5325\times10^{5}}{500}=613\,\mathrm{m^{2}\,s^{-2}}
Take the square root.
v_b=\sqrt{613}=24.76\,\mathrm{m\,s^{-1}}

v_b=24.76\,\mathrm{m\,s^{-1}}

E=1.5325\times10^{5}\,\mathrm{J},\qquad v_b=24.76\,\mathrm{m\,s^{-1}}

Book answer. 153\,250\,\mathrm{J}, 24.75\,\mathrm{m\,s^{-1}}

Note. The energy agrees with the book. The speed differs in the last digit: \sqrt{613}=24.7588, which rounds to 24.76 m/s. The book's 24.75 m/s appears to be truncated. The book says the cart 'weighs' 500 kg; 500 kg is its mass.

5.6

A car engine operates with 25 % efficiency and produces 50 kJ of useful work. How much fuel energy did the car consume?

Solution

Given

\eta = 25\,\% efficiency
W_{\text{out}} = 50\,\mathrm{kJ} useful work

To find

W_{\text{in}}, the fuel energy

Efficiency is useful output divided by input, so the input is the output divided by the efficiency.
\eta=\frac{W_{\text{out}}}{W_{\text{in}}}\quad\Rightarrow\quad W_{\text{in}}=\frac{W_{\text{out}}}{\eta}
Substitute the values, with \eta=0.25.
W_{\text{in}}=\frac{50\,\mathrm{kJ}}{0.25}=200\,\mathrm{kJ}

W_{\text{in}}=200\,\mathrm{kJ}

Book answer. 200\,\mathrm{kJ}

5.7

A refrigerator uses 600 W of electrical power and has an efficiency of 15 %. How much cooling power (useful energy) does the refrigerator produce in one hour?

Solution

Given

P = 600\,\mathrm{W} electrical power
\eta = 15\,\% efficiency
t = 1\,\mathrm{h}=3600\,\mathrm{s}

To find

the useful energy produced in one hour

Find the electrical energy supplied in one hour.
E_{\text{in}}=Pt=600\times3600=2.16\times10^{6}\,\mathrm{J}
The useful energy is the efficiency times the energy supplied.
E_{\text{out}}=\eta\,E_{\text{in}}=0.15\times2.16\times10^{6}=3.24\times10^{5}\,\mathrm{J}
As a check, the useful power is 0.15\times600\,\mathrm{W}=90\,\mathrm{W}, and 90\,\mathrm{W}\times3600\,\mathrm{s}=324\,\mathrm{kJ}.

E_{\text{out}}=324\,\mathrm{kJ}\ \text{in one hour}

Book answer. 324\,\mathrm{kJ}

Note. The book calls 324 kJ the 'cooling power'. Power is energy per second, so the cooling power is 90 W, and 324 kJ is the useful energy delivered in one hour.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 61 mark

Two bodies A and B of same masses have velocities in the ratio 2:1, their kinetic energies are in the ratio:

A2:1
B1:2
C4:1Answer
D1:4

Why

Kinetic energy is \tfrac12mv^{2}. For equal masses, \dfrac{K_A}{K_B}=\dfrac{v_A^{2}}{v_B^{2}}=\left(\dfrac{2}{1}\right)^{2}=4:1. Option A (2:1) forgets that the speed is squared.

2025 Annual I, Set B, Q.2 (iii)3 marks

What is meant by 'conservative' and 'Non-conservative' forces? Give one example each.

Solution

Conservative force (1.5 marks). A force is conservative if the work it does in moving a body between two points is independent of the path followed, or equivalently if the work done around a closed path is zero. The field of such a force is a conservative field. Examples: gravitational force and electric force.

Non-conservative force (1.5 marks). A force is non-conservative if the work done depends on the path followed, or if the work done around a closed path is not zero. Examples: friction, viscous drag and air resistance.

	Conservative force	Non-conservative force
Work between two points	independent of the path	depends on the path followed
Work around a closed path	zero	not zero
Examples	gravitational force, electric force	friction, viscous drag, air resistance

2026 Annual I, Set B, Q.2 (iv)3 marks

Differentiate between conservative and non-conservative forces with examples.

Solution

	Conservative force	Non-conservative force
Work between two points	independent of the path	depends on the path followed
Work around a closed path	zero	not zero
Examples	gravitational force, electric force	friction, viscous drag, air resistance

2025 Annual I, Set B, Q.4 OR6 marks

State work-energy theorem. Explain this theorem for resistive medium.

Solution

Work-energy theorem (3 marks)

Statement. The net work done on an object is equal to the change in its kinetic energy.

W_{net}=\text{K.E.}_f-\text{K.E.}_i=\Delta\text{K.E.}=\tfrac12mv_f^{2}-\tfrac12mv_i^{2}

Derivation Work-energy theorem

Aim. Show that the net work done on a body equals its change in kinetic energy.

Assumptions.

The net force F is constant and acts along the displacement d.

Work is force times displacement.
W_{net}=Fd
Newton's second law gives the force.
F=ma
The third equation of motion, solved for d.
v_f^{2}=v_i^{2}+2ad\ \Rightarrow\ d=\dfrac{v_f^{2}-v_i^{2}}{2a}
Substitute F and d; the acceleration a cancels.
W_{net}=ma\cdot\dfrac{v_f^{2}-v_i^{2}}{2a}=\tfrac12mv_f^{2}-\tfrac12mv_i^{2}

ResultW_{net}=\Delta\text{K.E.}

Positive net work increases the kinetic energy; negative net work decreases it.

Work-energy theorem in a resistive medium (3 marks)

A resistive force (friction, viscous drag) opposes the motion. When an applied force does work W on an object, part of it increases the object's kinetic energy and the rest is the work W_r done by the object against the resistive force:

W=\Delta\text{K.E.}+W_r\qquad\Rightarrow\qquad W_r=W-\Delta\text{K.E.}

The work against the resistive force takes energy away from the object. If the speed is constant, \Delta\text{K.E.}=0 and all the work is spent against resistance, W_r=W. This method works even when the forces are not constant, because only the work and the change in kinetic energy are needed. A parachute, which greatly increases the resistive force, makes a skydiver slow down.

2026 Annual I, Set B, Q.63+3 = 6 marks

a. Derive the expression for Kinetic energy. b. State work energy theorem in a resistive medium.

Solution

Kinetic energy (3 marks)

Derivation Kinetic energy from the equations of motion

Aim. Derive the kinetic energy of a body of mass m that starts from rest and reaches speed v under a constant force.

Assumptions.

The force F is constant and acts in the direction of the displacement.
The mass m is constant and no other force, such as friction, does work on the ball.

Figure 5.7: Work done transfers kinetic energy to a cricket ball.

Write the work done by the force over the displacement. The force and displacement are in the same direction, so \theta=0^\circ.
W=Fd\cos0^\circ=Fd
Newton's second law gives the force in terms of the acceleration a it produces.
F=ma
Substitute the force into the work, so that the work depends on m, a and d.
W=mad
Use the third equation of motion, which links the initial speed v_i, the final speed v_f, the acceleration and the displacement. Solve it for d.
v_f^2=v_i^2+2ad\quad\Rightarrow\quad d=\dfrac{v_f^2-v_i^2}{2a}
Substitute this d into W=mad. The acceleration cancels.
W=ma\times\dfrac{v_f^2-v_i^2}{2a}=\dfrac12 m\left(v_f^2-v_i^2\right)
The work is equal to the energy gained by the ball. Write K.E=\tfrac12 mv^2 for a body of speed v, so the gain is K.E_f-K.E_i.
W=K.E_f-K.E_i
The ball starts from rest, so v_i=0 and K.E_i=0. Write v_f=v.
W=\dfrac12 mv^2

ResultK.E=\tfrac12 mv^2

Since \mathrm{kg\,m^{2}\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}, the SI unit of kinetic energy is the joule. Although derived for a cricket ball, the formula holds for any moving body.

Work-energy theorem in a resistive medium (3 marks)

Statement. The net work done on an object is equal to the change in its kinetic energy, W_{net}=\Delta\text{K.E.}. In a resistive medium (friction, air resistance, viscous drag) a force opposes the motion, so only part of the work W done by the applied force increases the kinetic energy and the rest, W_r, is the work done by the object against the resistive force:

W=\Delta\text{K.E.}+W_r\qquad\Rightarrow\qquad W_r=W-\Delta\text{K.E.}

2025 Annual I, Set D, MCQ 51 mark

The bodies A and B of equal masses have velocities ratio 2:1, their K.E are in the ratio of:

A2:1
B1:2
C4:1Answer
D1:4

Why

Kinetic energy is \tfrac12mv^{2}. For equal masses, \dfrac{K_A}{K_B}=\dfrac{v_A^{2}}{v_B^{2}}=\left(\dfrac{2}{1}\right)^{2}=4:1. Option A (2:1) forgets that the speed is squared.

2026 Annual I, Set D, Q.2 (iv)1+1+1 = 3 marks

In a science experiment a 3.00\,\mathrm{kg} water rocket, is launched from ground. The rocket's total energy at the top of its flight is 2352\,\mathrm{J}. a. What was the rocket's launching speed? b. What height did the rocket went?

Solution

Given

m = 3.00\,\mathrm{kg} mass of the rocket
E = 2352\,\mathrm{J} total energy at the top
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) launching speed v_0
(b) height h

Energy at the top (1 mark). Only gravity does work on the rocket (air resistance is ignored), so the total energy K.E+P.E is conserved (\Delta K.E=-\Delta U). At the top of a vertical flight the rocket is momentarily at rest, so its energy there is all potential energy, mgh. At the ground, taking h=0, all of it is kinetic energy.
E=\tfrac12mv_0^{2}\ (\text{ground})=mgh\ (\text{top})=2352\,\mathrm{J}
(a) Launching speed (1 mark).
v_0=\sqrt{\dfrac{2E}{m}}=\sqrt{\dfrac{2\times2352}{3.00}}=\sqrt{1568}=39.6\,\mathrm{m\,s^{-1}}
(b) Height (1 mark).
h=\dfrac{E}{mg}=\dfrac{2352}{3.00\times9.8}=80\,\mathrm{m}

v_0=39.6\,\mathrm{m\,s^{-1}},\quad h=80\,\mathrm{m}

2025 Annual I, Set D, Q.2 (iv) OR3 marks

What is meant by work? How can work done be calculated graphically by a variable force?

Solution

Work (1 mark)

Work is the scalar (dot) product of the force and the displacement. Its SI unit is the joule (1\,\mathrm{J}=1\,\mathrm{N\,m}).

W=\vec F\cdot\vec d=Fd\cos\theta

Work from the force-displacement graph (2 marks)

Work done by a constant force

Constant force

Figure 5.3: Work done by a constant force is the area of the rectangle under the force-displacement graph.

Example: a constant force of 5\,\mathrm{N} acts along a displacement of 6\,\mathrm{m}. The area is the work done.
W=F\,d=5\,\mathrm{N}\times6\,\mathrm{m}=30\,\mathrm{J}

Work done by a variable force

W_{\text{Total}}=\sum_{i=1}^{n}F_i\cos\theta_i\,\Delta d_i

When the force increases uniformly with displacement, the graph is a straight line and the area under it is a triangle (Figure 5.4).

Variable force

Figure 5.4: Work done by a variable force is the area of the region under the force-displacement graph.

Example: the force rises uniformly from 0 to 6\,\mathrm{N} over 5\,\mathrm{m}. The area under the graph is a triangle, whose area is half the base times the height.
W=\tfrac12\times d\times F_{\max}
Substitute d=5\,\mathrm{m} and F_{\max}=6\,\mathrm{N}.
W=\tfrac12\times5\,\mathrm{m}\times6\,\mathrm{N}=15\,\mathrm{J}

In every case the work done is the area between the force-displacement graph and the displacement axis. The method works for any shape of graph, as the area is the sum of the strips.

2025 Annual I, Set D, Q.2 (v) OR3 marks

Differentiate between conservative and non-conservative forces. Give examples.

Solution

	Conservative force	Non-conservative force
Work between two points	independent of the path	depends on the path followed
Work around a closed path	zero	not zero
Examples	gravitational force, electric force	friction, viscous drag, air resistance

2025 Annual I, Set D, Q.2 (vi) OR3 marks

Derive the formula for Kinetic Energy [using the equations of motion]

Solution

Derivation Kinetic energy from the equations of motion

Aim. Derive the kinetic energy of a body of mass m that starts from rest and reaches speed v under a constant force.

Assumptions.

The force F is constant and acts in the direction of the displacement.
The mass m is constant and no other force, such as friction, does work on the ball.

Figure 5.7: Work done transfers kinetic energy to a cricket ball.

Write the work done by the force over the displacement. The force and displacement are in the same direction, so \theta=0^\circ.
W=Fd\cos0^\circ=Fd
Newton's second law gives the force in terms of the acceleration a it produces.
F=ma
Substitute the force into the work, so that the work depends on m, a and d.
W=mad
Use the third equation of motion, which links the initial speed v_i, the final speed v_f, the acceleration and the displacement. Solve it for d.
v_f^2=v_i^2+2ad\quad\Rightarrow\quad d=\dfrac{v_f^2-v_i^2}{2a}
Substitute this d into W=mad. The acceleration cancels.
W=ma\times\dfrac{v_f^2-v_i^2}{2a}=\dfrac12 m\left(v_f^2-v_i^2\right)
The work is equal to the energy gained by the ball. Write K.E=\tfrac12 mv^2 for a body of speed v, so the gain is K.E_f-K.E_i.
W=K.E_f-K.E_i
The ball starts from rest, so v_i=0 and K.E_i=0. Write v_f=v.
W=\dfrac12 mv^2

ResultK.E=\tfrac12 mv^2

Since \mathrm{kg\,m^{2}\,s^{-2}}=\mathrm{N\,m}=\mathrm{J}, the SI unit of kinetic energy is the joule. Although derived for a cricket ball, the formula holds for any moving body.

2026 Annual I, Set D, Q.5 OR6 marks

A ball of 100gm is thrown vertically upward at a speed of 25\,\mathrm{m/s}. If no energy is lost, determine the height it would reach. If the ball only rises to 25\,\mathrm{m}, calculate the work done against the air resistance. Also calculate the force of friction.

Solution

Given

m = 100\,\mathrm{g}=0.100\,\mathrm{kg} mass of the ball
v_i = 25\,\mathrm{m\,s^{-1}} launch speed
h' = 25\,\mathrm{m} height actually reached
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(a) the height h if no energy is lost
(b) the work W_r done against air resistance
(c) the force of friction (air resistance) F_r

The ball leaves the hand with kinetic energy, and at the highest point its speed is zero, so K.E_f=0.
K.E_i=\tfrac12mv_i^{2}=\tfrac12\times0.100\times(25)^{2}=31.25\,\mathrm{J}

(a) Height reached if no energy is lost.

No energy is lost, so all the initial kinetic energy becomes potential energy at the top.
\tfrac12mv_i^{2}=mgh
Solve for h (the mass cancels).
h=\dfrac{v_i^{2}}{2g}=\dfrac{(25)^{2}}{2\times9.8}=\dfrac{625}{19.6}=31.9\,\mathrm{m}

h=31.9\,\mathrm{m}

(b) Work done against the air resistance if the ball only rises to 25\,\mathrm{m}.

Gain in potential energy over the actual rise of 25\,\mathrm{m}.
P.E=mgh'=0.100\times9.8\times25=24.5\,\mathrm{J}
Work-energy theorem in a resistive medium: the kinetic energy supplied goes partly to potential energy and partly into the work W_r against air resistance. The ball ends at rest, so
K.E_i=mgh'+W_r\ \Rightarrow\ W_r=31.25-24.5=6.75\,\mathrm{J}

W_r=6.75\,\mathrm{J}

The resistive force acts over the whole 25\,\mathrm{m} of the rise, opposite to the motion, so W_r=F_rh'.
F_r=\dfrac{W_r}{h'}=\dfrac{6.75}{25}
Evaluate.
F_r=0.27\,\mathrm{N}

F_r=0.27\,\mathrm{N}

h=31.9\,\mathrm{m},\quad W_r=6.75\,\mathrm{J},\quad F_r=0.27\,\mathrm{N}

Exam focus

Derive the formula for kinetic energy, \tfrac12mv^{2}, using the equations of motion. (long)
Q.6 (a), 3 marks, in 2026 Set B and Q.2 (vi) OR in 2025 Set D; same as long question 1 of the book and an SLO.
Go to answer
State the work-energy theorem and explain it in a resistive medium (W=\Delta\text{K.E.}+W_r). (long)
Q.6 (b), 3 marks, in 2026 Set B and Q.4 OR (6 marks) in 2025 Set B; long question 3 of the book.
Go to answer
Differentiate between conservative and non-conservative forces, with examples. (short)
Q.2 (iv) in 2026 Set B, Q.2 (iii) in 2025 Set B and Q.2 (v) OR in 2025 Set D: three of the last four papers; short question 4 of the book.
Go to answer
Numericals on the resistive medium and conservation of energy: ball thrown upward, work against air resistance, F_r=W_r/h. (numerical)
Q.5 OR (6 marks) in 2026 Set D (ball of 100 g at 25\,\mathrm{m\,s^{-1}}) and Q.2 (iv) in 2026 Set D (water rocket, K.E+P.E conserved). Book numericals 3 to 5.
Go to answer
What is work? Find work by a constant and by a variable force from the force-displacement graph. (short)
Q.2 (iv) OR in 2025 Set D (not in 2026); long question 2 and an SLO.
Go to answer
Kinetic energy ratio MCQs: K\propto v^{2} (speed doubled or halved). (mcq)
MCQ 6 in 2025 Set B and MCQ 5 in 2025 Set D (the same question); not in 2026. Book exercise MCQs 5 to 7.
Go to answer
Efficiency of machines and energy losses in practical devices. (mcq)
Not in the 2025 or 2026 paper; book exercise MCQs 8 to 11 and numericals 6 and 7 are all on it.
Go to answer