NCERT Solutions Class 9 Science Chapter 7 In-text Questions — Work, Energy, and Simple Machines

NCERT Class 9 Science — Work, Energy, and Simple Machines, the seven Pause and Ponder boxes on pages 119, 121, 123, 126, 129, 132 and 135. All 13 questions solved.

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

The seven boxes follow the chapter through its whole argument: what work is in physics, then kinetic energy, then potential energy, then the conservation of the two together, and finally the simple machines that trade force against distance. Four relations carry all thirteen questions:

• Work $= F \times s$, the force multiplied by the displacement in the direction of the force. It is negative when the two are opposite, and zero when there is no displacement.
• Kinetic energy $= \dfrac12 mv^2$, which depends on the speed and nothing else.
• Potential energy $= mgh$ near the Earth’s surface, which depends on the height and nothing else.
• Mechanical advantage $= \dfrac{\text{load}}{\text{effort}}$, which for an inclined plane is $\dfrac{L}{h}$ and for a lever is the ratio of the arms.

Key insight. A machine never reduces the work; it only spreads it out. Lengthening a ramp or a lever arm lets you push with a smaller force, but you push through a correspondingly longer distance, and $F \times s$ comes to the same total. Questions 9 to 12 are four versions of that one trade, and Question 13 is what happens to the difference when friction is in the room.

Pause and Ponder, page 119. The box that follows Section 7.2, on positive and negative work.

Question 1

In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?

Solution. In everyday speech, holding a heavy barbell overhead is obviously hard work. In physics the word means something narrower:

$$W = F \times s$$

where $s$ is the displacement of the object in the direction of the force. The weightlifter certainly applies a large upward force. But while she holds the barbell steady, the barbell does not move at all: $s = 0$.

$$W = F \times 0 = 0$$

So no work is done on the barbell, however tiring the holding is. Her muscles are using up chemical energy all the while — that is what makes it exhausting — but none of it is going into the barbell.

Question 2

Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) — positive, negative or zero?

Solution. Friction always opposes relative motion, so as the stack of coins slides forward, the force of friction on it points backwards.

The displacement is forwards and the force is backwards: they are in opposite directions. By the chapter’s own rule, work done by a force whose direction is opposite to the displacement is negative.

That sign is not a technicality. Negative work means friction is taking energy out of the coins, which is exactly why they slow down and stop.

Pause and Ponder, page 121. A single question, after the section on energy.

Question 3

When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?

Solution. Follow the energy rather than the bicycle. The chemical energy in your food becomes muscular energy, and pedalling passes it to the machine and its rider.

Part of it becomes kinetic energy — the energy of the moving bicycle and yourself, most obviously while you are speeding up. On a flat road at steady speed the kinetic energy stops growing, yet you must keep pedalling, and the reason is where the rest goes.

The rest is dissipated as thermal energy: in friction at the chain, the bearings and the brakes, between the tyres and the road, and in pushing air out of the way. A little also leaves as sound. None of this is lost — it warms the machine and the air — but none of it is recoverable as motion, which is why the pedalling can never stop.

Pause and Ponder, page 123. The box that follows Section 7.4.1, on kinetic energy.

Question 4

Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?

Solution. Write the kinetic energies and set them equal, since that is what the question gives.

$$\frac12 m v_A^2 = \frac12 (4m) v_B^2$$

The $\tfrac12$ and the $m$ cancel from both sides:

$$v_A^2 = 4 v_B^2 \quad\Longrightarrow\quad \frac{v_A}{v_B} = 2$$

So A moves twice as fast as B. That the ratio is 2 and not 4 is the whole point: kinetic energy goes as the square of the speed, so quadrupling the mass only doubles the speed needed to match it.

Question 5

Does the kinetic energy of an object which moves with constant velocity change with its position?

Solution. Kinetic energy is $\tfrac12 mv^2$ — it depends on the mass and the speed, and on nothing else. Position does not appear in it anywhere.

If the velocity is constant then the speed is constant, so $\tfrac12 mv^2$ has the same value wherever the object happens to be. The kinetic energy does not change.

It is worth contrasting this with the next question: potential energy is the one that depends on position, and kinetic energy is the one that depends on motion.

Pause and Ponder, page 126. A single question, after Section 7.4.2 on potential energy.

Question 6

Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?

Solution. Near the Earth’s surface the potential energy is

$$U = mgh$$

and the only variable in it is $h$, the height above the chosen reference level. So the question becomes: does the height change?

Moving horizontally: no. Carrying an object across a level floor leaves $h$ exactly as it was, so $mgh$ is unchanged and the potential energy does not change — however far it is carried, and whatever its speed.

Raised vertically: yes. Now $h$ increases, so $U = mgh$ increases with it. Raising the object through a height $\Delta h$ increases its potential energy by $mg\,\Delta h$, which is precisely the work done against gravity in lifting it.

Pause and Ponder, page 129. The box that follows Section 7.4.3, on the conservation of mechanical energy.

Question 7

For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is $mgh$.

Solution. In Fig. 7.19 the object is dropped from rest at A, a height $h$ above the ground, and C is the ground itself. The chapter has already shown that the mechanical energy is $mgh$ at A and at the intermediate point B; what is left is the last point.

Potential energy at C. The ball is at ground level, so $h = 0$:

$$U = mg \times 0 = 0$$

Kinetic energy at C. The ball fell from rest through a height $h$ under gravity, so by the kinematic relation $v^2 = u^2 + 2gh$ with $u = 0$:

$$v^2 = 2gh$$

$$K = \frac12 m v^2 = \frac12 m (2gh) = mgh$$

Mechanical energy at C.

$$E = K + U = mgh + 0 = mgh$$

So the total is the same $mgh$ as at A and at B. What has happened between A and C is not a change in the total but a complete handover: all of the potential energy the ball started with has turned into kinetic energy.

Question 8

You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Solution. In Fig. 7.22 the ball is released at the top of the tall ramp and runs over a series of humps of falling height: A is the first hump, B the dip after it, then C, D and E, each lower than the one before.

At A, the top of the first hump, the ball is at its greatest height on that stretch and moving at its slowest: the potential energy is at a maximum and the kinetic energy at a minimum.

At B, the dip, the height is least and the ball is moving fastest: the potential energy is at a minimum and the kinetic energy at a maximum. The energy has simply changed form on the way down.

At C, the next hump, the ball climbs again, so the kinetic energy falls and the potential energy rises — but not to what it was at A, because C is a lower hump.

Why the humps get lower. If mechanical energy were perfectly conserved, the ball could rise to the same height every time, and equal humps would do. In a real track it cannot. Friction between the ball and the rail, and air resistance, do negative work on the ball at every moment, converting part of its mechanical energy into thermal energy and sound. That energy warms the track and the air and does not come back.

So the total mechanical energy falls steadily, and with it the greatest height the ball can still reach. Each hump must therefore be built lower than the last, or the ball would not get over it. Yes — it is exactly the energy lost to friction.

Pause and Ponder, page 132. The box that follows Section 7.6, on the inclined plane.

Question 9

Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?

Solution. (The figure reference points back to Chapter 4, which carries “Fig. 4.26: A mountain road” — it is a genuine cross-reference, not a misprint.)

A winding hill road is an inclined plane. For a vehicle to reach the top of a hill of height $h$, the work done against gravity is $mgh$ whichever route it takes — that is fixed by the height alone.

What the route changes is how that work is divided between force and distance. The mechanical advantage of an inclined plane is

$$\text{mechanical advantage} = \frac{L}{h}$$

where $L$ is the length of the slope. Winding around the hill makes $L$ much greater than the height $h$, so the ratio is large and the force needed to push the vehicle up at each moment is correspondingly small — small enough for an engine to supply, and for brakes to hold on the way down.

A road straight up the hill would have $L$ barely larger than $h$, a mechanical advantage close to 1, and would demand a force close to the full weight of the vehicle. The total work is the same either way; the winding road simply spreads it over a longer distance.

Question 10

To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.

Solution. The same trade, on a smaller scale. Reaching a floor at height $h$ means raising your own weight through $h$, so the work done against gravity, $mgh$, is the same on either ladder.

A vertical ladder has its length equal to that height, so at each step you must lift your whole weight straight up. An inclined ladder is longer than the height it spans, $L > h$, and part of your weight is taken by the rungs pushing along the slope. The mechanical advantage $L/h$ is greater than 1, so the force you have to exert at each step is smaller.

The price is that you climb a longer distance and take more steps. Less force, more distance, the same total work — which is what every simple machine does.

Pause and Ponder, page 135. The last box, after the section on levers.

Question 11

Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?

Solution. The spoon is being used as a lever, and it is worth naming its three parts. The rim of the can, where the spoon rests, is the fulcrum. The lid being prised up is the load. The hand pressing down on the far end of the handle supplies the effort. The fulcrum is between the two, so it is a class I lever.

Now compare the arms. The load arm — from the rim to the edge of the lid — is very short, a centimetre or two. The effort arm — from the rim to your hand at the end of the handle — is the whole length of the spoon. The mechanical advantage is the ratio of the two:

$$\text{mechanical advantage} = \frac{\text{effort arm}}{\text{load arm}}$$

which is large. So a gentle push at the handle produces a much larger force on the lid, enough to prise it off — while your hand moves a long way and the lid moves only a little.

Question 12

Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?

Solution. Scissors are a pair of class I levers sharing one pivot: the screw is the fulcrum, the hand at the handles supplies the effort, and the object being cut is the load.

The mechanical advantage is again the ratio of the arms:

$$\text{mechanical advantage} = \frac{\text{effort arm}}{\text{load arm}}$$

Moving the object closer to the screw makes the load arm shorter while the effort arm — the distance from the screw to your fingers — stays the same. The ratio therefore increases, so the same squeeze of the hand produces a larger force on the object.

That is exactly what a hard object needs. Place it out near the tips instead, where the load arm is long, and the mechanical advantage is small — fine for paper, not for a thick wire.

Question 13

Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.

Solution. A perpetual machine would have to run for ever on the energy it started with, so the question is whether that energy stays available.

In every real machine, surfaces rub and parts move through air. Friction and air resistance act opposite to the motion at every moment, so they do negative work on the moving parts — they continually take mechanical energy out of the machine.

That energy is not destroyed. It becomes thermal energy in the bearings, the axles and the surrounding air, and a little sound. But it is no longer mechanical energy, and nothing in the machine can gather it back up and turn it into motion again.

So the mechanical energy falls steadily. The machine slows, and when what remains is no longer enough to overcome friction, it stops. The only way to keep it going is to supply energy continuously from outside — fuel, electricity, a hand on a crank. A machine does not create energy; it only passes energy on, and always with something lost to friction on the way.

Common mistakes

• Saying the weightlifter does work because holding is tiring. The physics definition needs a displacement, and there is none. Her muscles are spending chemical energy, but no work is done on the barbell.
• Giving the ratio in Question 4 as 4:1. Kinetic energy goes as $v^2$, so four times the mass needs only twice the speed to match it. Forgetting the square is the commonest slip in this chapter.
• Writing that friction “does no work” on the sliding coins because it is only a resisting force. It does work, and the work is negative — which is precisely how the coins lose their kinetic energy.
• Claiming a machine reduces the work needed. It never does. A ramp or a lever reduces the force and increases the distance by the same factor; $F \times s$, and so the work, is unchanged.
• Explaining the falling roller-coaster humps as “the ball runs out of energy”. Energy is conserved overall — what runs out is the mechanical part of it, because friction has converted some into thermal energy the ball cannot use.
• Placing a hard object at the tips of the scissors. That lengthens the load arm and lowers the mechanical advantage, which is the opposite of what Question 12 is asking for.

Practise next

• The end-of-chapter questions, Revise, Reflect, Refine, where the same formulae appear with numbers — work done against friction, kinetic and potential energy of falling objects, and the mechanical advantage of levers.
• Chapter 6, How Forces Affect Motion — its in-text questions set up the forces that these questions now follow over a distance.