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?

Accepted Answer

No. While she holds the barbell steady, she does no work on it. Work done by a force is the force multiplied by the displacement of the object in the direction of the force: $W = F \times s$ She does apply an upward force on the barbell (equal to its weight), but while it is held steady the barbell does not move at all, so its displacement is zero: $W = F \times 0 = 0$ She did do work on the barbell while lifting it up (force upward, displacement upward). Once it is held still, no more work is done on it. Why does she still feel tired? Her muscles keep contracting to hold the weight up, and they use up chemical energy from food, most of which becomes heat inside her body. That is energy used inside her body — it is not work done on the barbell , which neither moves nor gains any energy.

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?

Accepted Answer

The work done by friction on the coins is negative. Friction always acts opposite to the direction of motion of the stack of coins. The coins move forward, while the force of friction on them points backward. When the force and the displacement are in opposite directions, the work done by that force is negative: $W_{\text{friction}} = -\,(\text{friction force}) \times (\text{distance moved})$ This negative work takes kinetic energy away from the coins. That is exactly why the stack slows down and finally stops on the rough surface — the lost kinetic energy appears mostly as heat in the coins and the surface.

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?

Accepted Answer

Your muscles get their energy from the chemical energy of the food you eat. As you ride on a flat road, this energy appears in these forms: 1. Kinetic energy of you and the bicycle — the energy of motion (including the turning pedals, chain and wheels). 2. Thermal energy (heat) — your body warms up and you sweat; friction in the chain, the wheel bearings, between the tyres and the road, and the push of the air against you all turn some energy into heat. 3. Sound energy — a small amount, such as the whirr of the chain and the noise of the tyres on the road. On a flat road your height does not change, so no gravitational potential energy is gained. To keep moving at a steady speed you must keep pedalling, because friction and air resistance keep converting your kinetic energy into heat.

Question 4

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

Accepted Answer

Let the speeds of A and B be $v_A$ and $v_B$. Use $K = \tfrac{1}{2}mv^2$ for each object. $K_A = \tfrac{1}{2}\,m\,v_A^2$ $K_B = \tfrac{1}{2}\,(4m)\,v_B^2$ The kinetic energies are equal, so: $\tfrac{1}{2}\,m\,v_A^2 = \tfrac{1}{2}\,(4m)\,v_B^2$ Cancel $\tfrac{1}{2}m$ from both sides: $v_A^2 = 4\,v_B^2$ $\frac{v_A^2}{v_B^2} = 4$ Take the square root of both sides (speeds are positive): $\frac{v_A}{v_B} = \sqrt{4} = 2$ So $v_A : v_B = 2 : 1$. The lighter object A moves twice as fast as B. Mass is 4 times larger for B, but speed enters as its square, so B needs only half the speed to carry the same kinetic energy.

Question 5

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

Accepted Answer

No. Kinetic energy depends only on the mass of the object and its speed: $K = \tfrac{1}{2}mv^2$ If the object moves with constant velocity , both its mass $m$ and its speed $v$ stay the same, so $\tfrac{1}{2}mv^2$ stays the same at every position along its path. Position does not appear anywhere in the formula for kinetic energy. This also fits the work-energy theorem: constant velocity means zero acceleration, so the net force on the object is zero. Zero net force does zero net work, so the kinetic energy cannot change.

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?

Accepted Answer

Near the Earth’s surface the gravitational potential energy is

$U = mgh$

where $h$ is the height above the ground. It depends only on the height, not on the horizontal position.

(a)Moving horizontally with constant velocity: the height $h$ does not change, so $mgh$ does not change. The potential energy stays the same. (Its kinetic energy also stays the same, since the velocity is constant.)

(b)Gradually raised vertically: the height $h$ increases, so $mgh$ increases. The potential energy increases — by $mg \times$ (the extra height). This increase comes from the work done by the force that lifts the object.

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 $mgh$.

Accepted Answer

In Fig. 7.19 the object of mass $m$ is dropped from rest from point A at height $h$. At time $t$ after release its speed and height are (from the book’s kinematic equations): $v = gt$ $h' = h - \tfrac{1}{2}gt^2$ Step 1 — the instant just before it hits the ground. At the ground its height is zero: $h' = 0$ $\tfrac{1}{2}gt^2 = h$ Step 2 — potential energy there. Height is zero, so $U = mg \times 0 = 0$ Step 3 — kinetic energy there. $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m(gt)^2$ $K = mg \times \left(\tfrac{1}{2}gt^2\right)$ The bracket is exactly $h$ (from Step 1), so $K = mgh$ Step 4 — mechanical energy. $E = K + U = mgh + 0 = mgh$ So just before it hits the ground, all the potential energy $mgh$ it had at A has turned into kinetic energy, and the total mechanical energy is still $mgh$ — the same as at A and at B. Mechanical energy is conserved during free fall (air resistance neglected). Speed at the ground: from $\tfrac{1}{2}mv^2 = mgh$ we get $v = \sqrt{2gh}$, the same result as the book’s Example 7.8 (child on a slide).

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?

Accepted Answer

The ball starts at rest at the highest point, so at the start it has maximum potential energy and zero kinetic energy. As it rolls, the two keep changing into each other:

Point	What happens to the energy
Going down from the start	Height falls, so potential energy falls; the ball speeds up, so kinetic energy rises.
A (a peak)	The ball climbs to A, so it slows down: kinetic energy changes back into potential energy. At A potential energy is high (a local maximum) and kinetic energy is low (a local minimum), but not zero — the ball must still be moving to go over the top.
B (the valley)	The ball rolls down from A to B. B is the lowest point between A and C, so here potential energy is least and kinetic energy is greatest; the ball moves fastest at B.
C (next peak)	Climbing from B to C, kinetic energy again changes into potential energy. C is lower than A, so at C the potential energy is less than at A.

Why are C, D, E lower and lower? If there were no friction, mechanical energy (potential + kinetic) would stay constant, and the ball could rise to almost its starting height again. But on a real track there is friction between the ball and the rails, and air resistance. These forces act opposite to the motion and do negative work on the ball all the time, turning some of its mechanical energy into heat (and a little sound).

So the ball’s total mechanical energy keeps decreasing as it travels. With less total energy, it can climb only to a smaller height each time. A peak that is as high as or higher than the previous one would never be crossed. That is why the designer makes each later peak (C, then D, then E) lower than the one before it. Yes — this is exactly because of the energy lost to friction.

Question 9

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

Accepted Answer

A road going up a hill is an inclined plane . For an inclined plane of length $L$ rising through a height $h$ (ignoring friction): $F' \times L = mgh$ $\text{mechanical advantage} = \frac{\text{load}}{\text{effort}} = \frac{mg}{F'} = \frac{L}{h}$ The height of the hill $h$ is fixed. A road that winds around in gentle slopes is much longer ($L$ is large), so $\tfrac{L}{h}$ is large and the force $F'$ needed to move a vehicle up is much smaller. • Vehicles (and people) need a smaller force from the engine at any moment, so they can climb without stalling or overheating. • The total work, $mgh$, is the same either way — it is just spread over a longer distance. • A gentle slope is also safer : vehicles can grip the road, brake and control their speed while coming down, which would be very hard on a steep road going straight up.

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.

Accepted Answer

An inclined ladder works like an inclined plane , a simple machine. On a vertical ladder you have to raise your whole body straight up: at each rung your arms and legs must supply an upward force equal to your full weight $mg$, and your arms must also hold you against the ladder. On an inclined ladder you reach the same height $h$ by moving a longer distance $L$ along the slope. Part of your weight is supported by the ladder itself (you lean on it), so the force you have to apply at each step is smaller: $F' = \frac{mgh}{L} < mg$ The work done to reach the floor, $mgh$, is the same in both cases. But on the inclined ladder it is done with a smaller force over a longer path, which feels much easier and is also safer, since you do not have to hang on with your arms.

Question 11

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

Accepted Answer

The spoon acts as a lever (a Class I lever): • Fulcrum: the rim of the can, where the spoon rests. • Load: the tight lid, pushed up by the tip of the spoon just inside the rim — a very short load arm. • Effort: your hand pressing down on the handle — a long effort arm. For a lever: $\text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}$ $\text{mechanical advantage} = \frac{\text{effort arm}}{\text{load arm}}$ Because the effort arm (the handle) is many times longer than the load arm (the tiny tip), the mechanical advantage is large. A small push on the handle produces a large upward force on the lid, enough to pry it loose. Your hand moves through a larger distance than the lid does, so no extra work is being created — the force is simply multiplied.

Question 12

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

Accepted Answer

A pair of scissors is a lever with the fulcrum at the screw (pivot). Your fingers apply the effort at the handles, and the object being cut is the load, pressed by the blades. $\text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}$ The effort arm (pivot to your fingers) stays the same. When you push the hard object closer to the pivot , the load arm becomes shorter. So the same effort produces a larger cutting force on the object: $\text{force on object} = \text{effort} \times \frac{\text{effort arm}}{\text{load arm}}$ A shorter load arm means a larger mechanical advantage (effort arm ÷ load arm), so a hard object such as thick cardboard or a wire is cut more easily near the pivot than near the tips of the blades.

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.

Accepted Answer

A machine cannot create energy; it can only change energy from one form to another or transfer it. In every real machine there are forces like friction at the axles and moving parts, and air resistance. These forces always act opposite to the motion, so they do negative work on the moving parts.

By the work-energy theorem, negative work means a decrease in energy. So, little by little, the machine’s mechanical (kinetic + potential) energy is converted into heat and sound, which spreads into the surroundings and cannot run the machine again.

$\text{energy at start} = \text{energy left} + \text{energy lost as heat and sound}$

Unless energy is supplied from outside (fuel, electricity, a push, falling water), the energy left keeps shrinking, so the machine slows down and finally stops. A perpetual motion machine would have to keep doing work forever without any input, which would mean creating energy from nothing — that goes against the conservation of energy. Wheels, weights or magnets only store and move energy around; they cannot make new energy.

Question 14

State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.

Accepted Answer

(i) False. Work = force × displacement in the direction of the force. If the object does not move, the displacement is zero and so the work done is zero, however large the force. (ii) True. The lifting force is upward and the bucket moves upward — force and displacement are in the same direction, so the work done is positive. (iii) True. Work done on an object appears as a change in its energy, so both are measured in the same unit, the joule (1 J = 1 N × 1 m). (iv) False. It is not moving, so its kinetic energy is zero. Because it is stretched (deformed), it has elastic potential energy . (v) True. For example, a bulb changes electrical energy into light and heat, and a falling object changes potential energy into kinetic energy.

Question 15

Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ______. (iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______. (v) Power is defined as the ______ at which work is done.

Accepted Answer

(i) Work done = force × displacement (in the direction of force). (ii) 1 newton. (1 J = 1 N × 1 m.) (iii) $\tfrac{1}{2}mv^2$ (iv) $mgh$ (v) Power is defined as the rate at which work is done.

Question 16

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.

Accepted Answer

Correct statements: (iii) and (iv). (i) ✗ Incorrect. Gravity still pulls the ball down with a force equal to its weight, $mg$, even at the top. The force is not zero. (ii) ✗ Incorrect. Since the force is $mg$, the acceleration is $g$ (about 10 m s −2 ) directed downward. Only the velocity is zero at that instant, not the acceleration — that is why the ball immediately starts coming down. (iii) ✓ Correct. At the highest point the ball is momentarily at rest ($v = 0$), so $K = \tfrac{1}{2}mv^2 = 0$. (iv) ✓ Correct. It is at its greatest height, so $U = mgh$ is the largest at this point. All the kinetic energy it was thrown with has changed into potential energy (ignoring air resistance). This assumes the ball is thrown straight up. If it were thrown at a slant, it would still be moving sideways at the top, so its kinetic energy would not be zero there.

Question 17

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.

Accepted Answer

Situation Energy transformation (i) Truck moving uphill Chemical energy of the fuel → kinetic energy of the truck + gravitational potential energy (it gains height), with some heat and sound. (ii) Unwinding of a watch spring Elastic potential energy of the wound spring → kinetic energy of the gears and hands. (iii) Photosynthesis in green leaves Light energy of the Sun → chemical energy stored in food (glucose). (iv) Water flowing from a dam Gravitational potential energy of the stored water → kinetic energy of the flowing water (in a power station, this then turns turbines and becomes electrical energy). (v) Burning of a matchstick Chemical energy → heat and light energy (the rubbing first turns a little mechanical energy into heat that starts the burning). (vi) Explosion of a fire cracker Chemical energy → heat, light, sound and kinetic energy of the flying pieces and gases. (vii) Speaking into a microphone Sound energy → electrical energy (electrical signal). (viii) A glowing electric bulb Electrical energy → light energy + heat (thermal) energy. (ix) A solar panel Light (solar) energy → electrical energy.

Question 18

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 m, acceleration due to gravity is g = 10 m s −2 , and student’s mass is m = 50 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?

Accepted Answer

Given: $m = 50\ \text{kg}$, $g = 10\ \text{m s}^{-2}$, $h = 72.5\ \text{m}$ (ground level taken as zero potential energy). (i) Lifted straight up in the elevator. $\Delta U = mgh = 50\ \text{kg} \times 10\ \text{m s}^{-2} \times 72.5\ \text{m}$ $\Delta U = 500\ \text{N} \times 72.5\ \text{m} = 36250\ \text{J}$ The gain in potential energy is 36250 J (about 36.3 kJ) . (ii) Climbing the stairs. The student goes along a different, longer path, but starts at the same ground level and ends at the same height of 72.5 m. The vertical rise $h$ is the same, so $\Delta U = mgh = 500\ \text{N} \times 72.5\ \text{m} = 36250\ \text{J}$ The gain is again 36250 J . (iii) Conclusion. The gain in gravitational potential energy depends only on the change in height between the starting and the final positions, not on the path taken to get there. Whether lifted straight up or by a winding staircase, the student gains the same $mgh$. (Climbing the stairs will make the student more tired, because muscles also produce a lot of heat — but the potential energy gained is the same.)

Question 19

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.

Accepted Answer

Let each floor be $h$ high and let the first lift take time $t$. (Both lifts start from the ground.) Energy. The energy needed equals the gain in potential energy $mgH$: $E_1 = mg(10h)$ $E_2 = mg(20h) = 2 \times mg(10h) = 2E_1$ So the second lift needs twice the energy , that is, an extra amount equal to $E_1$ (the energy for 10 more floors). Power. Power = energy ÷ time: $P_1 = \frac{E_1}{t}$ $P_2 = \frac{2E_1}{2t} = \frac{E_1}{t} = P_1$ No extra power is needed — the power is the same. Twice the work is done, but in twice the time, so the rate of doing work is unchanged. If instead the crane took the mass from the 10th floor on up to the 20th floor, it would raise it through 10 floors again: the same energy $E_1$ as the first lift, but in time $2t$, so only half the power would be needed.

Question 20

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.

Accepted Answer

Factors. The energy needed is the gain in potential energy of the flag: $E = mgh$ So it depends on (a) the mass (weight) of the flag ($mg$) and (b) the height $h$ through which it is raised (the height of the pole). In a real pulley, some extra energy is also needed to overcome friction in the pulley and to lift the rope. Slowly or quickly? The work done in raising the flag to the top is $mgh$ either way, because it depends only on the weight and the height, not on the time taken. Raising it slowly or quickly does not change the work done (ignoring friction and the small kinetic energy given to the flag while it moves). Doubling the speed. If the speed is doubled, the flag covers the same height in half the time. Power is work ÷ time: $P_1 = \frac{mgh}{t}$ $P_2 = \frac{mgh}{t/2} = 2 \times \frac{mgh}{t} = 2P_1$ The same work in half the time means the power required is doubled .

Question 21

A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 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.

Accepted Answer

All the energy from the fuel goes into the kinetic energy of the moving mass (scooter + riders), so the fuel used is proportional to the kinetic energy gained from rest.

Day 1: moving mass = 60 kg + 100 kg = 160 kg.

$K_1 = \tfrac{1}{2} \times 160\ \text{kg} \times v^2$

Day 2: moving mass = 60 kg + 100 kg + 40 kg = 200 kg.

$K_2 = \tfrac{1}{2} \times 200\ \text{kg} \times v^2$

The speed $v$ is the same, so $\tfrac{1}{2}v^2$ cancels in the ratio:

$\frac{K_1}{K_2} = \frac{160}{200} = \frac{4}{5}$

So the fuel used on the two days is in the ratio 4 : 5 (the second day uses 25% more fuel). The time interval being the same does not change the energy; it only means the power on Day 2 is also larger in the same ratio.

Question 22

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.

Accepted Answer

A seesaw is a lever with the fulcrum in the middle. It balances when

$\text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}$

Let the child’s weight be $W$, so the adult’s weight is $2W$. Let the adult sit at distance $d$ from the fulcrum and the child at distance $x$ on the other side:

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

$x = 2d$

So the child must sit twice as far from the fulcrum as the adult. For example, if the adult sits 1 m from the fulcrum, the child sits 2 m from it.

The balanced seesaw: the adult (twice as heavy) sits at distance d from the fulcrum; the child sits twice as far, at 2d. 2W × d = W × 2d.

The heavier person sits closer to the fulcrum and the lighter person farther away, so that weight × distance is equal on both sides.

Question 23

A ball of mass 2 kg is thrown up with a velocity of 20 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 m, how much work was done by air resistance (assume g = 10 m s −2 ).

Accepted Answer

(i) Gravity always pulls the ball downward . • Upward motion: displacement is upward, force is downward (opposite), so the work done by gravity is negative . The ball loses kinetic energy. • Downward motion: displacement is downward, in the same direction as gravity, so the work done by gravity is positive . The ball gains kinetic energy. (ii) Given $m = 2\ \text{kg}$, $u = 20\ \text{m s}^{-1}$, maximum height $h = 19.4\ \text{m}$, $g = 10\ \text{m s}^{-2}$. Step 1 — mechanical energy at the start (take the throwing point as zero height): $K_i = \tfrac{1}{2} \times 2\ \text{kg} \times (20\ \text{m s}^{-1})^2 = 1 \times 400 = 400\ \text{J}$ $E_i = K_i + U_i = 400\ \text{J} + 0 = 400\ \text{J}$ Step 2 — mechanical energy at the highest point (the ball is momentarily at rest, so $K_f = 0$): $U_f = mgh = 2\ \text{kg} \times 10\ \text{m s}^{-2} \times 19.4\ \text{m} = 20 \times 19.4 = 388\ \text{J}$ $E_f = 0 + 388\ \text{J} = 388\ \text{J}$ Step 3 — work done by air resistance. Gravity’s effect is already counted in the potential energy, so the only other force, air resistance, accounts for the change in mechanical energy: $W_{\text{air}} = E_f - E_i = 388\ \text{J} - 400\ \text{J} = -12\ \text{J}$ Air resistance did −12 J of work on the ball (negative, because it acts opposite to the motion). That is why the ball reached only 19.4 m instead of the 20 m it would reach without air: $h_{\text{no air}} = \frac{K_i}{mg} = \frac{400\ \text{J}}{20\ \text{N}} = 20\ \text{m}$

Question 24

A 10.0 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 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?

Accepted Answer

(i) Speed at 0 m. Using $K = \tfrac{1}{2}mv^2$ with $K = 180\ \text{J}$ and $m = 10\ \text{kg}$: $180 = \tfrac{1}{2} \times 10 \times v_0^2 = 5\,v_0^2$ $v_0^2 = 36\ \text{m}^2\,\text{s}^{-2}$ $v_0 = 6\ \text{m s}^{-1}$ (ii) Speed at 4 m. The force changes with position, so the work done is the area under the force–displacement graph . Split it into three parts: (a) From 0 to 1 m, a triangle: $\tfrac{1}{2} \times 1\ \text{m} \times 50\ \text{N} = 25\ \text{J}$ (b) From 1 m to 3 m, a rectangle: $2\ \text{m} \times 50\ \text{N} = 100\ \text{J}$ (c) From 3 m to 4 m, a triangle: $\tfrac{1}{2} \times 1\ \text{m} \times 50\ \text{N} = 25\ \text{J}$ $W = 25 + 100 + 25 = 150\ \text{J}$ Friction is negligible, so by the work-energy theorem all this work is added to the kinetic energy: $K_4 = 180\ \text{J} + 150\ \text{J} = 330\ \text{J}$ $330 = \tfrac{1}{2} \times 10 \times v_4^2 = 5\,v_4^2$ $v_4^2 = 66\ \text{m}^2\,\text{s}^{-2}$ $v_4 = \sqrt{66} \approx 8.1\ \text{m s}^{-1}$ Negative acceleration? No. The only horizontal force is the applied force, and it acts along the motion everywhere from 0 to 4 m (it is never below zero on the graph). So $a = \tfrac{F}{m}$ is always positive or zero, and the block keeps speeding up all the way. Between 3 m and 4 m the force gets smaller, so the acceleration decreases (to zero at 4 m) — but a decreasing acceleration is not a negative acceleration.

Question 25

The gravitational attraction on the surface of the Moon (lunar surface) is about $\tfrac{1}{6}$th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 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?

Accepted Answer

At the highest point all the kinetic energy has changed into potential energy (no air on the Moon, and we ignore it on the Earth too):

$\tfrac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

The ball is thrown with the same speed $v$ in both places, so $v^2$ is the same and the height is inversely proportional to $g$.

On the Earth:

$8\ \text{m} = \frac{v^2}{2g}$

On the Moon the acceleration is $\tfrac{g}{6}$:

$h_{\text{Moon}} = \frac{v^2}{2 \times \tfrac{g}{6}} = 6 \times \frac{v^2}{2g}$

$h_{\text{Moon}} = 6 \times 8\ \text{m} = 48\ \text{m}$

The ball will go up to about 48 m on the Moon — six times as high, because the Moon pulls it back with only one-sixth of the force.

Question 26

A 1000 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?

Accepted Answer

(i) A to B. The graph is a horizontal line at 35 m s −1 from 0 s to 1 s. So the car moves with a constant (uniform) speed of 35 m s −1 for 1 s, with no acceleration. This is the driver’s reaction time: the obstruction has been seen, but the brakes have not yet been applied. In this 1 s the car covers $35\ \text{m s}^{-1} \times 1\ \text{s} = 35\ \text{m}$ (ii) Kinetic energy at A. $m = 1000\ \text{kg}$, $v = 35\ \text{m s}^{-1}$: $K_A = \tfrac{1}{2} \times 1000\ \text{kg} \times (35\ \text{m s}^{-1})^2$ $K_A = 500 \times 1225 = 612500\ \text{J}$ So $K_A = 612500\ \text{J} \approx 6.1 \times 10^5\ \text{J}$. (The speed at B is the same, so $K_B$ is also 612500 J.) (iii) Work done by the brakes (B to C). At C the car is at rest, so its kinetic energy is zero. By the work-energy theorem: $W_{\text{brakes}} = K_C - K_B = 0 - 612500\ \text{J} = -612500\ \text{J}$ The brakes do −612500 J of work (negative, because the braking force acts opposite to the car’s motion). Cross-check with force × distance: from B to C the speed falls from 35 to 0 m s −1 in 2 s, so the deceleration is 17.5 m s −2 and the braking force is 1000 × 17.5 = 17500 N. The distance B to C is the area of the triangle under the graph, $\tfrac{1}{2} \times 2 \times 35 = 35\ \text{m}$. Work = −17500 N × 35 m = −612500 J — the same answer. (iv) The kinetic energy is transformed mainly into heat (thermal energy) — the brake pads and discs (or drums), the tyres and the road all get hot because of friction — and a little into sound (the squeal of the brakes and tyres).

Question 27

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

Accepted Answer

The track is frictionless, so the mechanical energy stays constant . At O the ball is at rest: $E = K_O + U_O = 0 + 30\ \text{J} = 30\ \text{J}$ At any point, $K = E - U = 30\ \text{J} - U$, and then $v = \sqrt{\tfrac{2K}{m}}$ with $m = 0.5\ \text{kg}$. At P ($U = 20\ \text{J}$): $K_P = 30 - 20 = 10\ \text{J}$ $\tfrac{1}{2} \times 0.5 \times v_P^2 = 10$ $v_P^2 = \frac{10 \times 2}{0.5} = 40\ \text{m}^2\,\text{s}^{-2}$ $v_P = \sqrt{40} = 2\sqrt{10} \approx 6.3\ \text{m s}^{-1}$ At Q ($U = 30\ \text{J}$): $K_Q = 30 - 30 = 0$ $v_Q = 0\ \text{m s}^{-1}$ The ball just comes to rest at Q (it reaches Q, stops for an instant and rolls back). At R ($U = 40\ \text{J}$): $K_R = 30 - 40 = -10\ \text{J}$ Kinetic energy $\left(\tfrac{1}{2}mv^2\right)$ can never be negative, so this is impossible: the ball can never reach R. It does not have enough energy (it has only 30 J, but R needs at least 40 J of potential energy). It turns back at Q, so there is no velocity at R to calculate. Between O and Q the track never rises above 30 J of potential energy (the hump in the middle is below 20 J), so the ball can indeed travel from O, past P, up to Q.

Question 28

A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 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 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 m s −2 .

Accepted Answer

Given: $m = 1.5\ \text{kg}$, $h = 10\ \text{m}$, $g = 10\ \text{m s}^{-2}$, coconut starts from rest. (i) Potential energy at the top: $U = mgh = 1.5\ \text{kg} \times 10\ \text{m s}^{-2} \times 10\ \text{m} = 150\ \text{J}$ Just before it hits the sand, all of this has become kinetic energy (air resistance neglected): $\tfrac{1}{2}mv^2 = mgh$ $v^2 = 2gh = 2 \times 10 \times 10 = 200\ \text{m}^2\,\text{s}^{-2}$ $v = \sqrt{200} = 10\sqrt{2} \approx 14.1\ \text{m s}^{-1}$ The coconut hits the sand at about 14.1 m s −1 . (ii) The coconut brings 150 J of kinetic energy into the sand. The sand’s resistive force (3000 N) acts opposite to its motion over the depth $d$, and this work uses up all 150 J: $3000\ \text{N} \times d = 150\ \text{J}$ $d = \frac{150\ \text{J}}{3000\ \text{N}} = 0.05\ \text{m}$ The depression is 0.05 m = 5 cm deep. Strictly, the coconut also drops by $d$ while it sinks, losing a tiny extra $mgd = 15\ \text{N} \times d$ of potential energy. Including it gives $d = 150 \div 2985 \approx 0.0503\ \text{m}$ — still about 5 cm. The question tells us to use only the coconut’s energy at impact, so 5 cm is the expected answer.

Work, Energy and Simple Machines

Pause and Ponder — In-text Questions

Revise, Reflect, Refine — End-of-Chapter Questions

Situation	Energy transformation
(i) Truck moving uphill	Chemical energy of the fuel → kinetic energy of the truck + gravitational potential energy (it gains height), with some heat and sound.
(ii) Unwinding of a watch spring	Elastic potential energy of the wound spring → kinetic energy of the gears and hands.
(iii) Photosynthesis in green leaves	Light energy of the Sun → chemical energy stored in food (glucose).
(iv) Water flowing from a dam	Gravitational potential energy of the stored water → kinetic energy of the flowing water (in a power station, this then turns turbines and becomes electrical energy).
(v) Burning of a matchstick	Chemical energy → heat and light energy (the rubbing first turns a little mechanical energy into heat that starts the burning).
(vi) Explosion of a fire cracker	Chemical energy → heat, light, sound and kinetic energy of the flying pieces and gases.
(vii) Speaking into a microphone	Sound energy → electrical energy (electrical signal).
(viii) A glowing electric bulb	Electrical energy → light energy + heat (thermal) energy.
(ix) A solar panel	Light (solar) energy → electrical energy.

Work, Energy and Simple Machines

Pause and Ponder — In-text Questions

Revise, Reflect, Refine — End-of-Chapter Questions

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