NCERT Solutions Class 9 Science Chapter 7 — Work, Energy, and Simple Machines

NCERT Class 9 Science — Work, Energy, and Simple Machines, Chapter 7, Revise, Reflect, Refine. All 15 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

These are the fifteen “Revise, Reflect, Refine” questions at the end of Chapter 7 of the new Exploration textbook (2026). Questions 1 to 4 check the definitions: what counts as work, the units, and how energy changes from one form to another. Questions 5 to 7 are about potential energy and power. Questions 8 and 10 to 15 use the work-energy theorem to find speeds, heights and the work done by forces such as air resistance, brakes and sand, and three of them (11, 13 and 14) are read off a graph. Question 9 is a seesaw, which is a lever. Every answer rests on three sets of results from the chapter:

• Work — $W = F \times s$, the force multiplied by the displacement in the direction of the force. It is negative when the displacement is opposite to the force, and when the force changes, the work is the area under the force-displacement graph.
• Energy — kinetic energy $K = \dfrac{1}{2}mv^2$ and potential energy $U = mgh$. The work-energy theorem says that the work done on an object equals the change in its energy, and when only gravity does work, the mechanical energy $K + U$ stays the same.
• Power and levers — power $P = \dfrac{W}{t}$, the rate of doing work; and a lever balances when effort $\times$ effort arm $=$ load $\times$ load arm.

Where a question says to take $g = 10\ \text{m s}^{-2}$ (Questions 5, 10 and 15), that value is used; none of the other questions needs a value of $g$.

Key insight. Work is the way energy is handed from one thing to another, so the work done on an object is exactly the change in its energy. That lets you skip the force-by-force detail and simply compare the energy at the start with the energy at the end. It is how Question 10 finds the work done by air resistance without knowing the force, how Question 11 gets a final speed from the area under a graph, and how Questions 12, 14 and 15 find heights and speeds. It is also why a pulley or a seesaw (Questions 7 and 9) can make a job easier but can never reduce the work to be done.

Question 1

State whether True or False.

Question 1 (i)

Work is said to be done when a force is applied, even if the object does not move.

Solution. In science, work is done only when the force moves the object: $W = F \times s$, where $s$ is the displacement in the direction of the force. If the object does not move, $s = 0$, and so $W = F \times 0 = 0$, however large the force. Pushing hard on a rigid wall can make you tired, because your muscles use up your body’s internal energy, but you do no work on the wall.

Question 1 (ii)

Lifting a bucket vertically upward results in positive work done on the bucket.

Solution. To lift the bucket you pull it upwards, and it moves upwards. The force and the displacement are in the same direction, so the work done by your force on the bucket is positive, and the bucket gains potential energy. (At the same time gravity, which pulls downwards, does negative work on the bucket; the statement is about the lifting force.)

Question 1 (iii)

The SI unit for both work and energy is joule (J).

Solution. From $W = F \times s$, one joule is the work done when a force of $1\ \text{N}$ moves an object through $1\ \text{m}$: $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$. The work-energy theorem says that the work done on an object equals the change in its energy, and two quantities can only be equal if they are measured in the same unit. So energy is also measured in joules.

Question 1 (iv)

A motionless stretched rubber band has kinetic energy.

Solution. Kinetic energy is the energy an object has because it is moving, $K = \dfrac{1}{2}mv^2$. A motionless band has $v = 0$, so $K = 0$. What it does have is potential energy, stored because it has been stretched out of its normal shape. Let go of it and that stored energy can shoot a stone from a slingshot or push a stack of coins.

Question 1 (v)

Energy can change from one form to another.

Solution. This happens all around us. A bulb changes electrical energy into light, a falling object’s potential energy becomes kinetic energy, and the chemical energy in food becomes mechanical energy in our muscles. Question 4 below is a list of such changes.

Question 2

Fill in the blanks.

Question 2 (i)

Work done = ______ × ______ (in the direction of force).

Solution. This is the definition of work, $W = F \times s$: the force multiplied by the displacement it produces in its own direction.

Question 2 (ii)

1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.

Solution. Putting $F = 1\ \text{N}$ and $s = 1\ \text{m}$ into $W = F \times s$ gives $W = 1\ \text{N} \times 1\ \text{m} = 1\ \text{J}$.

Question 2 (iii)

The expression for kinetic energy of a body of mass $m$ and velocity $v$ is ______.

Solution. The chapter finds this from the work-energy theorem. The work needed to speed an object up from $u$ to $v$ is $W = \dfrac{1}{2}m(v^2 – u^2)$. Starting from rest, $u = 0$, so the energy the object gains by moving is $\dfrac{1}{2}mv^2$.

Question 2 (iv)

The potential energy of an object of mass $m$ at a small height $h$ from the Earth’s surface is ______.

Solution. To lift the object slowly you must apply an upward force equal to its weight, $mg$, through the height $h$. The work done, $mg \times h$, is stored as potential energy. The height has to be small because $g$ stays the same only near the Earth’s surface.

Question 2 (v)

Power is defined as the ______ at which work is done.

Solution. Power tells us how quickly work is done: $P = \dfrac{W}{t}$. One watt is one joule of work done every second.

Question 3

When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?

• (i) The force acting on the ball is zero.
• (ii) The acceleration of the ball is zero.
• (iii) Its kinetic energy is zero.
• (iv) Its potential energy is maximum.

Solution. More than one statement is correct here. At its highest point the ball stops for an instant before it starts to fall back, so its velocity there is zero. Take each statement in turn.

(i) is wrong. Gravity does not switch off at the top: the ball’s weight, $mg$, still pulls it downwards. If there were no force at all, the ball would simply stay up there. (ii) is wrong for the same reason. The acceleration is $g$, downwards, all the way up and all the way down (ignoring air resistance). A velocity of zero does not mean an acceleration of zero; at the top the velocity is changing from upwards to downwards, which is exactly what an acceleration does. (iii) is right. With $v = 0$, $K = \dfrac{1}{2}m \times 0^2 = 0$. (iv) is right. The ball is at its greatest height, so $U = mgh$ is at its largest: all the kinetic energy it was thrown with has turned into potential energy.

Question 4

For each of the following situations, identify the energy transformation that takes place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, and (ix) a solar panel.

For each one, ask where the energy is stored or comes from at the start, and what it has become at the end. The forms used are the ones in the chapter: mechanical (kinetic and potential), chemical, thermal, light, sound and electrical energy.

Question 4 (i)

A truck moving uphill.

Solution. The engine burns fuel, which stores chemical energy. That energy makes the truck move (kinetic energy) and lifts it higher up the hill (potential energy). Some is also lost as heat and sound from the engine.

Question 4 (ii)

Unwinding of a watch spring.

Solution. Winding the watch twists the spring out of shape, and the work done is stored in it as potential energy. As the spring unwinds, it turns the gears and the hands, so the stored energy becomes kinetic energy.

Question 4 (iii)

Photosynthesis in green leaves.

Solution. Leaves take in sunlight and use it to make food (glucose), in which the energy is stored in chemical bonds.

Question 4 (iv)

Water flowing from a dam.

Solution. Water held high behind a dam has potential energy because of its height. As it flows down it speeds up, so the potential energy becomes kinetic energy. In a hydroelectric power station this moving water then turns turbines to generate electrical energy.

Question 4 (v)

Burning of a matchstick.

Solution. The chemicals in the match head and the wood store chemical energy. When the match burns, this energy is given out as heat and light.

Question 4 (vi)

Explosion of a fire cracker.

Solution. The powder in the cracker stores chemical energy, which is released all at once. It comes out as heat, a flash of light, a loud bang, and the kinetic energy of the pieces and hot gases that fly outwards.

Question 4 (vii)

Speaking into a microphone.

Solution. Your voice is sound, which is energy carried by vibrations of the air. The microphone turns these vibrations into an electrical signal.

Question 4 (viii)

A glowing electric bulb.

Solution. The electric current supplies electrical energy to the bulb. The bulb gives it out as light, and it also gets warm.

Question 4 (ix)

A solar panel.

Solution. A solar panel absorbs sunlight and produces an electric current.

Question 5

A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is $h = 72.5\ \text{m}$, acceleration due to gravity is $g = 10\ \text{m s}^{-2}$, and student’s mass is $m = 50\ \text{kg}$.

(i) Find the gain in the potential energy if the student is lifted straight up to the top. (ii) Find the gain in the potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about the dependence of the potential energy on the path taken?

Question 5 (i)

Solution. The elevator lifts the student slowly, so it pushes upwards with a force equal to the student’s weight, $mg$, through the height $h$. By the work-energy theorem this work is stored as potential energy, $U = mgh$:

$$U = mgh = 50\ \text{kg} \times 10\ \text{m s}^{-2} \times 72.5\ \text{m} = 36250\ \text{J}$$

Question 5 (ii)

Solution. On the stairs the student rises a little with each step and also moves forwards. Moving forwards is at right angles to gravity, so it does no work against gravity and adds nothing to the potential energy. Only the rises count, and all the rises together add up to the height of the building, $72.5\ \text{m}$. The student ends at the same height with the same mass, so:

$$U = mgh = 50 \times 10 \times 72.5 = 36250\ \text{J}$$

Question 5 (iii)

Solution. The two paths are very different, one straight up and one winding up the staircase, but the gain in potential energy is the same. $U = mgh$ contains only the mass, $g$ and the height; it has no term for the route taken.

Question 6

A crane lifts a mass $m$ to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.

Solution. Both lifts start from the ground. Let each floor be $h$ high, so the 10th floor is $10h$ above the ground and the 20th floor is $20h$ above it. The energy the crane must supply is the potential energy the mass gains, $mg \times \text{height}$:

$$E_1 = mg \times 10h = 10\,mgh \qquad\qquad E_2 = mg \times 20h = 20\,mgh = 2E_1$$

So the second lift needs twice the energy, that is, as much again as the first. Power is energy divided by time. If the first lift takes a time $t$, the second takes $2t$:

$$P_1 = \frac{E_1}{t} \qquad\qquad P_2 = \frac{E_2}{2t} = \frac{2E_1}{2t} = \frac{E_1}{t} = P_1$$

Doubling the energy and doubling the time cancel each other, so the power stays the same.

Question 7

Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.

Solution. Raising the flag means doing work against gravity, and that work is stored as the flag’s potential energy, $mgh$. So the energy needed depends on the mass (weight) of the flag and on the height of the pole it is raised through. In a real pulley some extra work is also done against friction. The pulley itself saves no work: a fixed pulley only changes the direction of the force, so that you can pull down instead of lifting up.

The work done is $mgh$ whichever way the flag is raised. The formula has no time in it, so raising the flag slowly or quickly does not change the work done (leaving aside the small effects of friction and air resistance).

Power is different: $P = \dfrac{W}{t}$. If the flag is raised at double the speed, it covers the same height in half the time. The same work in half the time means

$$P_{\text{new}} = \frac{W}{t/2} = 2 \times \frac{W}{t} = 2P$$

Question 8

A man of mass $60\ \text{kg}$ rides a scooter of mass $100\ \text{kg}$. He accelerates the scooter to a velocity $v$. The next day, his son with a mass of $40\ \text{kg}$ joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.

Solution. With no losses, all the energy from the fuel becomes kinetic energy. By the work-energy theorem, the energy supplied equals the kinetic energy gained, $\dfrac{1}{2}Mv^2$, where $M$ is the total mass that is set moving: the scooter and everyone on it. So the fuel used is proportional to the kinetic energy.

On the first day the moving mass is the man and the scooter, $M_1 = 60 + 100 = 160\ \text{kg}$. On the second day the son is added, $M_2 = 60 + 100 + 40 = 200\ \text{kg}$. The final speed $v$ is the same, so

$$\frac{\text{fuel on day 1}}{\text{fuel on day 2}} = \frac{\tfrac{1}{2} \times 160 \times v^2}{\tfrac{1}{2} \times 200 \times v^2} = \frac{160}{200} = \frac{4}{5}$$

The time taken does not come into the fuel used. It would matter for the power: the engine has to work harder on the second day to supply the extra energy in the same time.

Question 9

On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.

Solution. A seesaw is a lever with its fulcrum in the middle. Like the beam balance in the chapter, it balances when effort $\times$ effort arm $=$ load $\times$ load arm. That is, the weight on each side multiplied by its distance from the fulcrum must be the same on both sides.

Let the child’s weight be $W$, so the adult’s weight is $2W$. If the adult sits at a distance $d$ from the fulcrum and the child at a distance $x$, then for balance

$$2W \times d = W \times x \qquad\Rightarrow\qquad x = 2d$$

The heavier adult must sit nearer the fulcrum, at half the child’s distance. For example, if the adult sits $1\ \text{m}$ from the fulcrum, the child sits $2\ \text{m}$ from it on the other side.

Question 10

A ball of mass $2\ \text{kg}$ is thrown up with a velocity of $20\ \text{m s}^{-1}$. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of $19.4\ \text{m}$, how much work was done by air resistance (assume $g = 10\ \text{m s}^{-2}$).

Question 10 (i)

Solution. Gravity always pulls the ball downwards. On the way up the ball moves upwards, opposite to the force, so the work done by gravity is negative, and the ball slows down and loses kinetic energy. On the way down the ball moves downwards, in the same direction as the force, so the work done by gravity is positive, and the ball speeds up.

Question 10 (ii)

Solution. Compare the ball’s mechanical energy at the start with its mechanical energy at the top. Measure heights from the point where it is thrown.

At the start: $K = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 2 \times 20^2 = 400\ \text{J}$, and $U = 0$. Total $= 400\ \text{J}$.

At the top the ball is momentarily at rest, so $K = 0$, and $U = mgh = 2 \times 10 \times 19.4 = 388\ \text{J}$. Total $= 388\ \text{J}$.

The mechanical energy has fallen by $12\ \text{J}$. Gravity’s part is already counted in the potential energy, so the only other force acting, air resistance, must account for the loss. By the work-energy theorem,

$$\text{work done by air resistance} = 388\ \text{J} – 400\ \text{J} = -12\ \text{J}$$

It is negative because air resistance acts against the motion. As a check, with no air resistance all $400\ \text{J}$ would become potential energy and the ball would rise to $h = \dfrac{400}{2 \times 10} = 20\ \text{m}$. It falls $0.6\ \text{m}$ short, and $mg \times 0.6 = 2 \times 10 \times 0.6 = 12\ \text{J}$.

Question 11

A $10.0\ \text{kg}$ block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at $0\ \text{m}$ till $4\ \text{m}$. If the block had a kinetic energy of $180\ \text{J}$ when it was at $0\ \text{m}$, find the block’s speed (i) at $0\ \text{m}$, and (ii) at $4\ \text{m}$. Does the block have negative acceleration in any portion of its motion?

The graph (Fig. 7.37, redrawn with the area under it shaded) shows the force rising steadily from $0$ to $50\ \text{N}$ over the first metre, staying at $50\ \text{N}$ from $1\ \text{m}$ to $3\ \text{m}$, and falling steadily back to $0$ at $4\ \text{m}$.

Question 11 (i)

Solution. The speed at $0\ \text{m}$ comes from the kinetic energy there, $K = \dfrac{1}{2}mv^2$, rearranged for $v$:

$$v^2 = \frac{2K}{m} = \frac{2 \times 180}{10} = 36 \qquad\Rightarrow\qquad v = 6\ \text{m s}^{-1}$$

Question 11 (ii)

Solution. The force is not constant, so the work cannot be found from a single $F \times s$. Instead, the work done is the area under the force-displacement graph between $0\ \text{m}$ and $4\ \text{m}$. Split it into three simple shapes:

$$\begin{aligned} 0 \text{ to } 1\ \text{m (triangle)}: &\quad \tfrac{1}{2} \times 1\ \text{m} \times 50\ \text{N} = 25\ \text{J} \\ 1 \text{ to } 3\ \text{m (rectangle)}: &\quad 2\ \text{m} \times 50\ \text{N} = 100\ \text{J} \\ 3 \text{ to } 4\ \text{m (triangle)}: &\quad \tfrac{1}{2} \times 1\ \text{m} \times 50\ \text{N} = 25\ \text{J} \end{aligned}$$

So the force does $25 + 100 + 25 = 150\ \text{J}$ of work. It is positive, because the force acts in the direction of motion. Friction is negligible, and the block’s weight and the floor’s upward push act at right angles to the motion, so they do no work. By the work-energy theorem the block’s kinetic energy therefore rises by $150\ \text{J}$:

$$K_{4\ \text{m}} = 180\ \text{J} + 150\ \text{J} = 330\ \text{J}$$

$$v^2 = \frac{2K}{m} = \frac{2 \times 330}{10} = 66 \qquad\Rightarrow\qquad v = \sqrt{66} \approx 8.12\ \text{m s}^{-1}$$

Question 11 (last part)

Solution. The acceleration is caused by the net force, $a = F/m$, and it points the same way as that force. Here the only horizontal force is the applied force, and it acts in the direction of motion everywhere from $0\ \text{m}$ to $4\ \text{m}$; it is never negative on the graph. Between $3\ \text{m}$ and $4\ \text{m}$ the force gets smaller, but it is still pushing forwards, so the block keeps speeding up, only more gently. The acceleration becomes zero at $4\ \text{m}$, where the force becomes zero, but it is never negative.

Question 12

The gravitational attraction on the surface of the Moon (lunar surface) is about $\dfrac{1}{6}$th of that on the surface of the Earth. An astronaut can throw a ball up to a height of $8\ \text{m}$ from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?

Solution. Use conservation of mechanical energy. The ball leaves the hand with speed $u$, and at its highest point it is momentarily at rest, so all its kinetic energy has become potential energy (air resistance is ignored, and the Moon has no air anyway):

$$\frac{1}{2}mu^2 = mgh \qquad\Rightarrow\qquad h = \frac{u^2}{2g}$$

The mass cancels, and for the same throwing speed $u$ the height is inversely proportional to $g$. On the Moon $g$ is one-sixth as large, so the ball rises six times as high:

$$h_{\text{Moon}} = \frac{u^2}{2 \times (g/6)} = 6 \times \frac{u^2}{2g} = 6 \times h_{\text{Earth}} = 6 \times 8\ \text{m} = 48\ \text{m}$$

Question 13

A $1000\ \text{kg}$ car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38.

(i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?

Question 13 (i)

Solution. From A to B the graph is a horizontal line at $35\ \text{m s}^{-1}$, from $0$ to $1\ \text{s}$. The speed does not change, so the car keeps moving at a constant $35\ \text{m s}^{-1}$. This is the time the driver takes to react: the obstruction has been seen but the brakes are not yet on. In that one second the car covers $35 \times 1 = 35\ \text{m}$.

Question 13 (ii)

Solution. At A the speed is $35\ \text{m s}^{-1}$:

$$K = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 35^2 = \frac{1}{2} \times 1000 \times 1225 = 612500\ \text{J}$$

Question 13 (iii)

Solution. At B the speed is still $35\ \text{m s}^{-1}$, so the kinetic energy is still $612500\ \text{J}$; at C the car has stopped and its kinetic energy is zero. By the work-energy theorem, the work done by the brakes equals the change in kinetic energy:

$$W = K_C – K_B = 0 – 612500\ \text{J} = -612500\ \text{J}$$

The work is negative because the braking force acts backwards while the car moves forwards. As a check from the graph, the area under BC gives the braking distance, $\dfrac{1}{2} \times 2\ \text{s} \times 35\ \text{m s}^{-1} = 35\ \text{m}$. The speed falls steadily, so the braking force is constant, and it must be $612500 \div 35 = 17500\ \text{N}$; then $W = 17500\ \text{N} \times (-35\ \text{m}) = -612500\ \text{J}$.

Question 13 (iv)

Solution. The brakes stop the car by friction: the brake pads rub against the wheels, and the tyres grip the road. Rubbing makes things hot, which is why brakes get hot after a hard stop. So the car’s kinetic energy becomes thermal energy (heat) in the brakes, the tyres and the road, with a little sound (the squeal of the brakes). None of it is stored as potential energy: the car stays on the same level road, so its height does not change, and the chapter notes that work done against friction does not store energy the way lifting an object does.

Note on the book’s answer. The answer key prints “potential energy” for this part. That cannot be right. The car’s height does not change, so its gravitational potential energy does not change, and the braking force is friction, which stores no energy. The kinetic energy becomes thermal energy (heat) and some sound.

Question 14

The potential energy-displacement graph of a $0.5\ \text{kg}$ ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is $0\ \text{m s}^{-1}$ and potential energy is $30\ \text{J}$. Calculate the velocity of the ball at P, Q and R.

Solution. The track is frictionless, so the ball’s mechanical energy, $K + U$, stays the same all along it. At O, $K = 0$ (the ball is at rest) and $U = 30\ \text{J}$, so the total is $30\ \text{J}$. At any other point, the kinetic energy is whatever is left over after the potential energy is taken away, $K = 30\ \text{J} – U$, and then $v^2 = \dfrac{2K}{m}$ with $m = 0.5\ \text{kg}$.

O sits on top of a small hump, so the ball runs down to the right into the first valley. It easily gets over the middle hump, which is only about $19\ \text{J}$ high, and then climbs the slope on the right.

At P, $U = 20\ \text{J}$, so $K = 30 – 20 = 10\ \text{J}$ and

$$v_P^2 = \frac{2 \times 10}{0.5} = 40 \qquad\Rightarrow\qquad v_P = \sqrt{40} = 2\sqrt{10} \approx 6.3\ \text{m s}^{-1}$$

At Q, $U = 30\ \text{J}$, so $K = 30 – 30 = 0$ and $v_Q = 0$. The ball stops there for an instant and then rolls back.

At R, $U = 40\ \text{J}$. That would need $K = 30 – 40 = -10\ \text{J}$, but kinetic energy, $\dfrac{1}{2}mv^2$, can never be negative. The ball simply does not have enough energy to climb to R. It turns back at Q and keeps rolling to and fro between O and Q.

Question 15

A coconut of mass $1.5\ \text{kg}$ falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is $10\ \text{m}$. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is $3000\ \text{N}$ and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume $g = 10\ \text{m s}^{-2}$.

Question 15 (i)

Solution. At the top of the tree the coconut is at rest with potential energy

$$U = mgh = 1.5 \times 10 \times 10 = 150\ \text{J}$$

As it falls (ignoring air resistance) this potential energy turns into kinetic energy, so just before it hits the sand $\dfrac{1}{2}mv^2 = 150\ \text{J}$:

$$v^2 = \frac{2 \times 150}{1.5} = 200 \qquad\Rightarrow\qquad v = \sqrt{200} = 10\sqrt{2} \approx 14.1\ \text{m s}^{-1}$$

This is the same as $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10}$, the result of the chapter’s slide example; the mass cancels out.

Question 15 (ii)

Solution. The coconut reaches the sand with $150\ \text{J}$ of kinetic energy and ends at rest, so it loses all $150\ \text{J}$. The sand does this by pushing up with $3000\ \text{N}$ while the coconut moves down through the depth $d$, so the sand does negative work, $-3000\ \text{N} \times d$. By the work-energy theorem:

$$-3000 \times d = 0 – 150 \qquad\Rightarrow\qquad d = \frac{150}{3000} = 0.05\ \text{m} = 5\ \text{cm}$$

Strictly, the coconut’s own weight, $15\ \text{N}$, keeps pulling it down while it sinks, which makes the depression very slightly deeper: $d = \dfrac{150}{3000 – 15} = 0.0503\ \text{m}$. To the accuracy of the data this is still $0.05\ \text{m}$.

Common mistakes

• Question 1(i): taking effort for work. Pushing a wall hard feels like work, so students mark the statement true. In science, work needs a displacement in the direction of the force; with $s = 0$ the work done on the wall is zero, however tired you feel.
• Question 3: thinking the force or acceleration is zero at the top. Because the ball stops for an instant, students assume nothing is acting on it. But its weight never switches off, and the acceleration stays $g$ downwards. Only the velocity, and so the kinetic energy, is zero there.
• Question 5(ii): expecting more potential energy by the longer path. The staircase is a longer route, so students expect a larger answer. Potential energy depends only on the height gained, $U = mgh$, and the height is the same $72.5\ \text{m}$ either way.
• Question 6: doubling the power along with the energy. The energy does double, but so does the time, and power is energy divided by time. The two doublings cancel, so the power is unchanged.
• Question 8: leaving out the scooter. Using only the riders’ masses gives $60 : 100$. The fuel has to set the whole load moving — man, son and scooter — so the masses to compare are $160\ \text{kg}$ and $200\ \text{kg}$, giving $4 : 5$.
• Question 11: reading a falling force as slowing down. Between $3\ \text{m}$ and $4\ \text{m}$ the graph slopes down, and students say the block decelerates there. A smaller forward force still speeds the block up; only a backward force could slow it. Another slip is adding the $150\ \text{J}$ to the speed instead of to the kinetic energy.
• Question 13(iv): saying “potential energy”. This is what the book’s key prints, but the car does not rise, and friction in the brakes stores no energy. The kinetic energy becomes heat (and a little sound), which is why brakes get hot.
• Question 14: finding a speed at R. Using $v = \sqrt{2(U_R – 30)/m}$ gives a number, but it answers the wrong question. The kinetic energy at R would be $30 – 40 = -10\ \text{J}$, which is impossible, so the ball turns back at Q and never reaches R.

Practise next

• The in-text Pause and Ponder questions of this chapter, thirteen across seven boxes, which take the chapter in order: what work means in physics, then kinetic and potential energy separately, then their conservation, then the simple machines that trade force against distance.
• Chapter 6, How Forces Affect Motion — the forces behind the work done here: Newton’s laws of motion, friction, and the weight $mg$ used in every potential-energy calculation.
• Chapter 4, Describing Motion Around Us — speed-time graphs and the equations of motion, which Question 13’s graph and the formula $K = \frac{1}{2}mv^2$ are built on.
• Chapter 10, Sound Waves: Characteristics and Applications — the next physics chapter, where energy is carried from place to place by vibrations.