Work, Energy, and Simple Machines

(i) What will be the magnitude of velocity of the child at the bottom of the blue slide?

(ii) Will two children of different masses reach the bottom of the same slide with the same velocity?

(iii) Which of the slides will result in the largest magnitude of velocity for the child at its bottom?

Answer

As the child slides down, the potential energy at the top gets converted into kinetic energy at the bottom (neglecting friction).

So,

$\dfrac{1}{2}$mv² = mgh

v = $\sqrt{2gh}$

(i) The magnitude of velocity of the child at the bottom of the blue slide is v = $\sqrt{2gh}$, where h is the height of the blue slide. Its value depends only on the height of the slide.

(ii) Yes. Since the velocity v = $\sqrt{2gh}$ depends only on the height h and not on the mass, two children of different masses will reach the bottom of the same slide with the same velocity.

(iii) Since v = $\sqrt{2gh}$, the velocity is largest for the slide with the greatest height. The slide that starts from the highest point will give the child the largest magnitude of velocity at its bottom.

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?

Answer

No, she is not doing any work on the barbell while holding it steady.

Work done = Force × Displacement

While holding the barbell steady, there is no displacement of the barbell (s = 0).

∴ W = F × 0 = 0

Hence, the work done on the barbell is zero, even though she may feel tired because her muscles use up the internal energy of her body.

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

Answer

The work done by friction on the stack of coins is negative.

The force of friction acts in a direction opposite to the direction of motion (displacement) of the coins. Since the force and the displacement are in opposite directions, the work done by friction is negative.

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

Answer

As you ride, the muscular energy supplied by your muscles appears mainly in the following forms:

(i) Kinetic energy — of the moving bicycle and the rider.

(ii) Thermal energy — produced due to friction in the moving parts of the bicycle, friction with the road and air resistance, and within the body.

(iii) Sound energy — a small amount of sound produced as the bicycle moves.

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?

Answer

Given,

Mass of A = m

Mass of B = 4m

Kinetic energy of A = Kinetic energy of B

We know,

Kinetic energy = $\dfrac{1}{2}$mv²

So,

$\dfrac{1}{2}$ × m × v_A² = $\dfrac{1}{2}$ × 4m × v_B²

v_A² = 4v_B²

$\dfrac{v_A^2}{v_B^2}$ = 4

$\dfrac{v_A}{v_B}$ = 2

Hence, the ratio of the magnitude of velocities of A and B is 2 : 1.

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

Answer

No. The kinetic energy of an object depends only on its mass and velocity.

Kinetic energy = $\dfrac{1}{2}$mv²

Since the object moves with a constant velocity, its kinetic energy remains the same and does not change with its position.

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?

Answer

The potential energy of an object near the surface of the Earth is given by U = mgh, which depends on its height h above the Earth's surface.

When the object moves with constant velocity in the horizontal direction, its height h does not change. Hence, its potential energy does not change.

When the object is gradually raised in the vertical direction, its height h increases. Hence, its potential energy increases.

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.

Answer

The ball is dropped from a height h at point A with initial velocity u = 0. Just before it hits the ground (at point C), its height becomes zero.

Let v be the velocity of the ball just before it hits the ground.

Using the equation of motion,

v² = u² + 2gh

v² = 0 + 2gh = 2gh

At this position (just before hitting the ground):

Kinetic energy = $\dfrac{1}{2}$mv² = $\dfrac{1}{2}$ × m × 2gh = mgh

Potential energy = mg × 0 = 0 (since height = 0)

Mechanical energy = Kinetic energy + Potential energy

Mechanical energy = mgh + 0 = mgh

Hence, the mechanical energy of the ball just before it hits the ground is mgh, which is the same as its initial mechanical energy at point A. This shows that the mechanical energy is conserved.

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?

Answer

The ball is released from the highest point A.

At A (highest point): The ball is at the greatest height and is momentarily at rest. So, its potential energy is maximum and its kinetic energy is zero (minimum).

At B (a lower point): As the ball rolls down from A to B, its height decreases. So, the potential energy decreases and gets converted into kinetic energy, which increases.

At C (the next peak, lower than A): As the ball rises again to C, kinetic energy gets converted back into potential energy. So, the kinetic energy decreases and the potential energy increases. However, since C is lower than A, the potential energy at C is less than that at A.

The subsequent points such as C, D and E usually have lower heights compared to the previous ones because some of the mechanical energy is continuously lost in overcoming friction and air resistance, and gets converted mainly into thermal energy and sound energy. As a result, the ball cannot rise back to the same height it started from. So, yes, the lowering of the heights is due to the energy lost to friction.

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

Answer

A road on a hill acts as an inclined plane. For a road that winds around in gentle slopes, the length L of the road is much larger than the height h to be climbed.

Mechanical advantage of an inclined plane = $\dfrac{L}{h}$

A longer road with a gentle slope has a larger value of $\dfrac{L}{h}$, which means a much smaller force (effort) is required to climb the hill. Although the vehicle has to travel a longer distance, the force needed at any moment is reduced, making it easier and safer for vehicles to climb. Going straight up would require a very large force, which the vehicle's engine may not be able to provide.

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

Answer

An inclined ladder acts as an inclined plane. To reach the same height h, the length L of an inclined ladder is greater than the height itself.

Mechanical advantage of an inclined plane = $\dfrac{L}{h}$

Since L is greater than h, the mechanical advantage is greater than 1, so a smaller force (effort) is needed to climb the inclined ladder. In the case of a vertical ladder, we have to lift our entire weight straight up, which requires a larger force. Hence, climbing an inclined ladder is easier, though we have to cover a larger distance.

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

Answer

When we open the lid of a can using a spoon, the spoon acts as a lever. The edge of the can acts as the fulcrum, the lid is the load, and the force applied at the other end of the spoon is the effort.

Since the effort arm (the part of the spoon we hold) is much longer than the load arm, a small effort applied over the longer arm produces a much larger force on the lid.

effort × effort arm = load × load arm

As the effort arm is large, only a small effort is needed to overcome the large load. Hence, it becomes easier to open the lid using a spoon.

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

Answer

A pair of scissors works as a lever with the pivot acting as the fulcrum.

For a lever,

effort × effort arm = load × load arm

When we push the hard object closer to the fulcrum, the load arm becomes smaller. For the same effort applied at the handles (long effort arm), a smaller load arm allows a much larger force to act on the object. This larger cutting force makes it easier to cut a hard object.

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.

Answer

All real machines eventually slow down and stop because of friction and air resistance.

While a machine works, some work has to be done against friction in its moving parts and against air resistance. This work continuously converts a part of the machine's mechanical energy into thermal energy and sound energy, which dissipates into the surroundings and cannot be recovered.

Since a machine cannot create energy on its own — it can only use the energy supplied to it — the mechanical energy keeps decreasing as it is lost to friction. If no external energy (fuel or electricity) is supplied to make up for this loss, the machine eventually runs out of usable energy and stops. This is why a perpetual motion machine, which would run forever without any energy input, is impossible to build.

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.

Answer

(i) False. Work is done only when the force produces a displacement of the object. If the object does not move, no work is done.

(ii) True. While lifting a bucket upward, the applied force and the displacement are both in the upward direction, so the work done is positive.

(iii) True. The SI unit of both work and energy is the joule (J).

(iv) False. A motionless stretched rubber band has potential energy (due to its deformed shape), not kinetic energy.

(v) True. Energy can be transformed from one form to another.

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.

Answer

(i) Work done = Force × Displacement (in the direction of force).

(ii) 1 joule of work is done when a force of 1 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 $\dfrac{1}{2}$mv².

(iv) The potential energy of an object of mass m at a small height h from the Earth's surface is mgh.

(v) Power is defined as the rate at which work is done.

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

1. The force acting on the ball is zero.
2. The acceleration of the ball is zero.
3. Its kinetic energy is zero.
4. Its potential energy is maximum.

Answer

The correct statements are 3 and 4.

Reason — At the highest point, the velocity of the ball becomes zero, so its kinetic energy is zero. Since the ball is at its maximum height, its potential energy is maximum.

However, the force of gravity (mg) continues to act on the ball even at the highest point, so the force is not zero. The acceleration is also not zero; it remains equal to g, the acceleration due to gravity, directed downward.

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.

Answer

(i) A truck moving uphill: Chemical energy (of fuel) ⟶ Kinetic energy + Potential energy

(ii) Unwinding of a watch spring: Potential energy (stored in spring) ⟶ Kinetic energy

(iii) Photosynthesis in green leaves: Light energy (solar) ⟶ Chemical energy

(iv) Water flowing from a dam: Potential energy ⟶ Kinetic energy

(v) Burning of a matchstick: Chemical energy ⟶ Thermal energy + Light energy

(vi) Explosion of a fire cracker: Chemical energy ⟶ Thermal energy + Light energy + Sound energy + Kinetic energy

(vii) Speaking into a microphone: Sound energy ⟶ Electrical energy

(viii) A glowing electric bulb: Electrical energy ⟶ Light energy + Thermal energy

(ix) A solar panel: Light energy (solar) ⟶ Electrical energy

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?

Answer

Given,

Mass (m) = 50 kg

Height (h) = 72.5 m

Acceleration due to gravity (g) = 10 m s^–2

We know,

Gain in potential energy = mgh

(i) When the student is lifted straight up to the top,

Gain in potential energy = mgh = 50 × 10 × 72.5 = 36250 J

Hence, the gain in potential energy = 36250 J.

(ii) When the student climbs the stairs to the same top, the height gained is the same (h = 72.5 m).

Gain in potential energy = mgh = 50 × 10 × 72.5 = 36250 J

Hence, the gain in potential energy = 36250 J.

(iii) We conclude that the gain in potential energy is the same in both cases. Hence, the potential energy depends only on the vertical height gained and is independent of the path taken.

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.

Answer

Let the height of each floor be h₀ and the time taken to reach the 10th floor be t.

Energy:

Energy to reach the 10th floor, E₁ = mg × (10h₀)

Energy to reach the 20th floor, E₂ = mg × (20h₀)

$\dfrac{E_2}{E_1}$ = $\dfrac{20h_0}{10h_0}$ = 2

So, E₂ = 2E₁

Hence, twice as much energy is required (the energy required is doubled).

Power:

Power to reach the 10th floor, P₁ = $\dfrac{E_1}{t}$ = $\dfrac{mg \times 10h_0}{t}$

Power to reach the 20th floor (in double the time, 2t), P₂ = $\dfrac{E_2}{2t}$ = $\dfrac{mg \times 20h_0}{2t}$ = $\dfrac{mg \times 10h_0}{t}$

So, P₂ = P₁

Hence, the power required is the same in both cases (no extra power is needed).

Thus, lifting the mass to the 20th floor requires double the energy but the same power.

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.

Answer

The energy (work) required to raise the flag is given by W = mgh.

So, the energy required depends on two factors:

(i) The weight of the flag (mg, i.e., its mass), and

(ii) The height of the flagpole (h).

(A fixed pulley only changes the direction of the effort; its mechanical advantage is 1, so it does not reduce the work done.)

Raising slowly or quickly: The work done depends only on the weight of the flag and the height raised, not on the time taken. So, raising the flag slowly or quickly does not change the amount of work done.

Doubling the speed: Power = $\dfrac{\text{Work done}}{\text{Time}}$

The work done remains the same. If the speed is doubled, the time taken becomes half. Since power is inversely proportional to time, the power requirement becomes double when the speed is doubled.

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.

Answer

The energy supplied by the fuel goes entirely into the kinetic energy of the scooter (with its riders).

Kinetic energy = $\dfrac{1}{2}$Mv²

Since the fuel used is proportional to the energy (and hence to the total mass for the same speed),

Day 1:

Total mass, M₁ = mass of man + mass of scooter = 60 + 100 = 160 kg

Energy, E₁ = $\dfrac{1}{2}$ × 160 × v²

Day 2:

Total mass, M₂ = mass of man + mass of scooter + mass of son = 60 + 100 + 40 = 200 kg

Energy, E₂ = $\dfrac{1}{2}$ × 200 × v²

Ratio of fuel used = $\dfrac{E_1}{E_2}$ = $\dfrac{160}{200}$ = $\dfrac{4}{5}$

Hence, the ratio of the fuel used on the two days is 4 : 5.

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.

Answer

A seesaw works as a lever with the fulcrum at its centre. For the seesaw to be balanced,

effort × effort arm = load × load arm

i.e., (weight of child) × (distance of child) = (weight of adult) × (distance of adult)

Let the weight of the child be W and the weight of the adult be 2W. Let the child sit at a distance d_child and the adult at a distance d_adult from the fulcrum.

W × d_child = 2W × d_adult

d_child = 2 × d_adult

So, the child must sit at twice the distance of the adult from the fulcrum. For example, if the adult sits at 1 m from the fulcrum, the child must sit at 2 m from the fulcrum to balance the seesaw.

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).

Answer

Given,

Mass (m) = 2 kg

Initial velocity (u) = 20 m s^–1

Acceleration due to gravity (g) = 10 m s^–2

(i) During upward motion, the force of gravity acts downward while the displacement is upward (opposite directions). Hence, the work done by gravity is negative.

During downward motion, the force of gravity and the displacement are both downward (same direction). Hence, the work done by gravity is positive.

(ii) Initial kinetic energy of the ball,

K = $\dfrac{1}{2}$mu² = $\dfrac{1}{2}$ × 2 × (20)² = 400 J

At the maximum height (19.4 m), the ball comes to rest, so its kinetic energy = 0.

Change in kinetic energy = Final KE – Initial KE = 0 – 400 = –400 J

Work done by gravity during the rise = –mgh = –(2 × 10 × 19.4) = –388 J

Using the work-energy theorem,

Total work done = Change in kinetic energy

Work done by gravity + Work done by air resistance = Change in kinetic energy

(–388) + W_air = –400

W_air = –400 + 388 = –12 J

Hence, the work done by air resistance = –12 J.

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?

Answer

Given,

Mass (m) = 10.0 kg

Kinetic energy at 0 m = 180 J

(i) Speed at 0 m:

Kinetic energy = $\dfrac{1}{2}$mv²

180 = $\dfrac{1}{2}$ × 10 × v²

v² = $\dfrac{180 \times 2}{10}$ = 36

v = 6 m s^–1

Hence, the speed of the block at 0 m is 6 m s^–1.

(ii) Speed at 4 m:

The work done by the variable force on the block is equal to the area under the force-displacement graph (Fig. 7.37) between 0 m and 4 m.

Area under the graph

= area of left triangle + area of rectangle + area of right triangle

= $\dfrac{1}{2}$ × 1 × 50 + 2 × 50 + $\dfrac{1}{2}$ × 1 × 50

= 25 + 100 + 25 = 150 J

Using the work-energy theorem,

Kinetic energy at 4 m = Kinetic energy at 0 m + Work done

Kinetic energy at 4 m = 180 + 150 = 330 J

Now,

$\dfrac{1}{2}$mv² = 330

$\dfrac{1}{2}$ × 10 × v² = 330

v² = 66

v = $\sqrt{66}$ ≈ 8.1 m s^–1

Hence, the speed of the block at 4 m is about 8.1 m s^–1.

Negative acceleration: The applied force is never opposite to the direction of motion of the block, and friction is negligible. The acceleration is positive wherever the applied force is positive and becomes zero only where the force is zero. Hence, the block does not have negative acceleration in any portion of its motion.

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 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?

Answer

Given,

Height reached on Earth, h_E = 8 m

Gravitational attraction on the Moon, g_M = $\dfrac{1}{6}$g_E

The ball is thrown with the same upward velocity u on both the Earth and the Moon. At the maximum height, the velocity becomes zero.

Using v² = u² – 2gh, with v = 0,

h = $\dfrac{u^2}{2g}$

Since u is the same in both cases, the height h is inversely proportional to g.

$\dfrac{h_M}{h_E}$ = $\dfrac{g_E}{g_M}$ = $\dfrac{g_E}{\frac{1}{6}g_E}$ = 6

h_M = 6 × h_E = 6 × 8 = 48 m

Hence, the ball will travel up to a height of 48 m from the surface of the Moon.

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?

Answer

Given,

Mass of the car (m) = 1000 kg

From the graph (Fig. 7.38), the speed of the car between A and B is constant at 35 m s^–1.

(i) Between positions A and B, the speed of the car remains constant at 35 m s^–1. This is the time before the driver applies the brakes (the reaction time), during which the car continues to move uniformly.

(ii) Kinetic energy of the car at A,

K = $\dfrac{1}{2}$mv² = $\dfrac{1}{2}$ × 1000 × (35)²

= $\dfrac{1}{2}$ × 1000 × 1225 = 612500 J

Hence, the kinetic energy of the car at A = 612500 J.

(iii) Between B and C, the brakes bring the car to a complete stop, so the final kinetic energy = 0.

Work done by the brakes = Final KE – Initial KE = 0 – 612500 = –612500 J

Hence, the work done by the brakes = –612500 J (negative, since the braking force opposes the motion).

(iv) The kinetic energy of the car transforms mainly into thermal energy in the brakes and tyres, and a small amount into sound energy.

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.

Answer

Given,

Mass (m) = 0.5 kg

At O, velocity = 0 m s^–1 and potential energy = 30 J

So, kinetic energy at O = 0 J

Total mechanical energy = Kinetic energy + Potential energy = 0 + 30 = 30 J

Since the track is frictionless, the total mechanical energy (30 J) is conserved at every point.

At any point, Kinetic energy = Total mechanical energy – Potential energy

At P (potential energy = 20 J):

Kinetic energy = 30 – 20 = 10 J

$\dfrac{1}{2}$mv² = 10

$\dfrac{1}{2}$ × 0.5 × v² = 10

v² = 40

v = $\sqrt{40}$ = $2\sqrt{10}$ ≈ 6.3 m s^–1

At Q (potential energy = 30 J):

Kinetic energy = 30 – 30 = 0 J

v = 0 m s^–1

At Q, the ball momentarily comes to rest and turns back.

At R (potential energy = 40 J):

Kinetic energy = 30 – 40 = –10 J

Since kinetic energy cannot be negative, the ball cannot reach point R. The potential energy at R (40 J) is greater than the total mechanical energy of the ball (30 J), so the ball turns back before reaching R.

Hence, the velocity at P is about 6.3 m s^–1, at Q is 0 m s^–1, and the ball does not reach R.

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.

Answer

Given,

Mass (m) = 1.5 kg

Height of tree (h) = 10 m

Acceleration due to gravity (g) = 10 m s^–2

(i) The potential energy of the coconut at the top gets converted into kinetic energy just before it hits the sand.

Using v² = u² + 2gh, with u = 0,

v² = 0 + 2 × 10 × 10 = 200

v = $\sqrt{200}$ = $10\sqrt{2}$ ≈ 14.14 m s^–1

Hence, the velocity of the coconut just before it hits the sand is about 14.14 m s^–1.

(ii) Energy of the coconut just before impact = its kinetic energy = potential energy lost

Energy = mgh = 1.5 × 10 × 10 = 150 J

This energy is used in doing work against the resistive force of the sand to create the depression.

Work done against the resistive force = Resistive force × depth of depression

Let the depth of the depression be d.

3000 × d = 150

d = $\dfrac{150}{3000}$ = 0.05 m

Hence, the depth of the depression made by the coconut in the sand is 0.05 m (i.e., 5 cm).