7.1

Example — Positive and Negative Work

Question

While exercising, a girl lifts a dumbbell and slowly lowers it down. Identify when the girl does positive work on the dumbbell and when she does negative work on it.

Answer

The girl applies a force equal to the weight of dumbbell to lift it up. When she moves the dumbbell up, the force is in the direction of displacement, so she does positive work on it.

When moving the dumbbell down, the force she applies to hold it is in a direction opposite to the displacement, so she does negative work on it.

↑ Positive Work

Force and displacement are in the same direction.

↓ Negative Work

Force and displacement are in opposite directions.

7.2

Example — Work Done by a Goalkeeper

Question

While saving a goal, a goalkeeper’s hand moved back by 15 cm as she stopped a ball while applying a force of 200 N. How much work did the goalkeeper do on the ball in stopping it?

Answer

The goalkeeper applied a force opposite to the motion of the ball, so she did negative work on the ball.

The displacement should be taken as negative because the ball moves in a direction opposite to the direction of the applied force.

Calculation

Work done = Force × Displacement of ball in the direction of force

= 200 N × (−0.15 m)

= −30 J

✓ Work done by the goalkeeper = −30 J

“`html Chapter 7: Work, Energy, and Simple Machines – Solved Examples

7.1 Example 7.1

Question

While exercising, a girl lifts a dumbbell and slowly lowers it down. Identify when the girl does positive work on the dumbbell and when she does negative work on it.

Answer

When she moves the dumbbell upward, the applied force is in the direction of displacement. Therefore, she does positive work.
When she moves the dumbbell downward, the force applied by her to control the dumbbell is opposite to its displacement. Therefore, she does negative work.

7.2 Example 7.2

Question

While saving a goal, a goalkeeper’s hand moved back by 15 cm as she stopped a ball while applying a force of 200 N. How much work did the goalkeeper do on the ball in stopping it?

Answer

The goalkeeper applied a force opposite to the motion of the ball. Therefore, the work done is negative.
Displacement: s = −15 cm = −0.15 m
Work done: W = F × s = 200 × (−0.15)
W = −30 J

7.3 Example 7.3

Question

In a game of carrom, a player played the shot to pocket the black coin. Identify who does work, and the changes in energy that occur at each collision.

Answer

The moving striker collides with the white coin and applies a force in the direction of its displacement. Thus, it does positive work on the white coin and increases its energy.
By Newton’s third law, the white coin applies an opposite force on the striker and does negative work on it, decreasing its energy.
Similarly, the white coin does positive work on the black coin, increasing its energy, while the black coin does negative work on the white coin.

7.4 Example 7.4

Question

If the velocity of a vehicle doubles in magnitude, what will its kinetic energy be compared to its original value?

Answer

Let the mass of the vehicle be m and its initial velocity be v.
Initial kinetic energy: K = ½mv²
New velocity = 2v.
New kinetic energy: K’ = ½m(2v)² = 4 × ½mv² = 4K
Therefore, the new kinetic energy is 4 times the original kinetic energy.

7.5 Example 7.5

Question

In one of their fastest deliveries, an Indian cricketer bowled a cricket ball with an approximate mass of 0.2 kg at a velocity of about 154.8 km h⁻¹. Calculate the kinetic energy of the ball at the time of its delivery.

Answer

Mass: m = 0.2 kg
Velocity: v = 154.8 km h⁻¹
Converting velocity: v = 154.8 ÷ 3.6 = 43 m s⁻¹
Kinetic energy: K = ½mv²
= ½ × 0.2 × 43²
= 184.9 J

7.6 Example 7.6

Question

A jet aircraft of mass 15000 kg lands on the deck of an aircraft carrier. To stop the aircraft within the short length of the deck a hook on the aircraft’s tail is caught in a wire stretched across the deck. The wire exerts an approximately constant backward force of 367500 N and stops the jet within 100 m. What was the velocity of the aircraft just before the wire caught the hook?

Answer

Mass: m = 15000 kg
Force exerted by the wire: F = 367500 N
Displacement: s = −100 m
Work done by the wire: W = F × s
= 367500 × (−100) = −36,750,000 J
Since the aircraft finally stops: K_f = 0.
By the work-energy theorem: W = K_f − K_i
Therefore: −36,750,000 = 0 − ½ × 15000 × v²
Solving: v² = 4900
v = 70 m s⁻¹
In km h⁻¹: 70 × 3.6 = 252 km h⁻¹

7.7 Example 7.7

Question

After taking a catch, a fielder threw the cricket ball of mass 200 g high up in the air about 10 m above the ground in celebration. How much potential energy does the ball have when the ball reaches its maximum height? Assume g = 10 m s⁻².

Answer

Mass: m = 200 g = 0.2 kg
Height: h = 10 m
Acceleration due to gravity: g = 10 m s⁻²
Potential energy: U = mgh
= 0.2 × 10 × 10
= 20 J

7.8 Example 7.8

Question

What will be the magnitude of velocity of the child on reaching the bottom of the slide of height h?

Answer

At the top, the child possesses gravitational potential energy: PE = mgh
At the bottom, this potential energy is converted into kinetic energy, assuming friction is negligible: KE = ½mv²
By conservation of mechanical energy: mgh = ½mv²
Cancelling m: gh = ½v²
Therefore: v = √(2gh)
Thus, the velocity depends on the height h and does not depend on the mass of the child.

7.9 Example 7.9

Question

A truck of mass 10000 kg is moving at 72 km h⁻¹ when its brakes fail. The driver steers it onto an escape ramp inclined at 30°, where the truck comes to a rest. If the sand exerts a force of 50000 N opposite to the truck’s motion, what is the minimum length of the ramp to be able to stop such a truck? Take g = 10 m s⁻².

Hint: For a 30° incline, the truck rises 1 m vertically for every 2 m it travels along the ramp.

Answer

Initial velocity: 72 km h⁻¹ = 20 m s⁻¹
Initial kinetic energy: K = ½mv² = ½ × 10000 × 20² = 2,000,000 J
Let the length of the ramp be d. From the given condition, the vertical height gained is: h = d / 2
Increase in potential energy: PE = mgh = 10000 × 10 × d/2 = 50000d
Work done by sand: W_sand = −50000d
Applying energy conservation: 2,000,000 = 50000d + 50000d
Therefore: d = 20 m
Minimum ramp length = 20 m

7.10 Example 7.10

Question

A weightlifter lifts a 75 kg mass by 2 m in 5 seconds. How much power would she require for this task?

Answer

Work done: W = mgh = 75 × 10 × 2 = 1500 J
Power: P = W/t = 1500/5 = 300 W

7.11 Example 7.11

Question

A car of mass 1000 kg starts from rest and reaches a speed of 72 km h⁻¹ in 10 seconds. Calculate the power of the engine required to achieve this start.

Answer

Initial velocity: u = 0 m s⁻¹
Final velocity: v = 72/3.6 = 20 m s⁻¹
Initial kinetic energy: 0 J
Final kinetic energy: K = ½mv² = ½ × 1000 × 20² = 200,000 J
Work done by the engine: 200,000 J
Power: P = W/t = 200,000/10 = 20,000 W

7.12 Example 7.12

Question

A person uses an inclined ramp to raise an object over a step 30 cm high. The ramp has a width of 40 cm. What is the mechanical advantage of the ramp that helps the person achieve the task?

Answer

Height: AB = 30 cm
Width: BC = 40 cm
Length of ramp: AC = √(30² + 40²) = √2500 = 50 cm
Mechanical advantage of the inclined plane: MA = L/h = 50/30 ≈ 1.67

7.13 Example 7.13

Question

For a seesaw having four seats A, B, D, E and fulcrum at C, AC = EC = 2 m and BC = DC = 1 m. On which seats should children of masses 15 kg and 30 kg sit to make the seesaw balanced?

Answer

Let the 15 kg child sit at seat A, which is 2 m from the fulcrum.
For balance, the moments about the fulcrum must be equal: 15 × 2 = 30 × L
Therefore: L = 30/30 = 1 m
Seat D is 1 m from the fulcrum on the opposite side. Therefore, the 30 kg child should sit on seat D.

Quick Revision

Work: W = F × s • Kinetic Energy: KE = ½mv² • Potential Energy: PE = mgh • Power: P = W/t

↑ Back to Top

“`

Chapter 7: Work, Energy, and Simple Machines – True/False Exercises

Chapter 7: Work, Energy, and Simple Machines

Revise, Reflect, Refine — Exercise Question 1 (True or False with Reasons)

Statement (i)

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

False

Reason: According to the scientific definition of work, if there is no displacement of the object, no work is done regardless of the force being applied[cite: 3, 22].

Statement (ii)

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

True

Reason: When an object is lifted, the applied upward force and the vertical displacement are in the same direction, which results in positive work[cite: 3, 4, 5, 22].

Statement (iii)

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

True

Reason: Both mechanical work and energy share the same SI unit, which is the joule (J), defined as 1 N × 1 m[cite: 3, 6, 22].

Statement (iv)

A motionless stretched rubber band has kinetic energy.

False

Reason: A motionless stretched rubber band possesses potential energy due to its deformation or shape, not kinetic energy[cite: 9, 22].

Statement (v)

Energy can change from one form to another.

True

Reason: Energy is versatile and can freely transform from one form to another, such as potential energy converting into kinetic energy during free fall[cite: 6, 12, 13, 22].

“`html Chapter 7: Work, Energy, and Simple Machines – Fill in the Blanks

Statement (i)

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

Correct Answer

Force × Displacement

Explanation: Work done by a constant force is equal to the product of the force and the displacement of the object in the direction of the force.

W = F × s

Statement (ii)

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

Correct Answer

Explanation: One joule is the amount of work done when a force of 1 newton produces a displacement of 1 metre in the direction of the force.

1 J = 1 N × 1 m

Statement (iii)

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

Correct Answer

½mv²

Explanation: Kinetic energy is the energy possessed by an object due to its motion. It depends on the mass and the square of the velocity of the object.

KE = ½mv²

Statement (iv)

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

Correct Answer

mgh

Explanation: The gravitational potential energy of an object of mass m at a height h above the Earth’s surface is given by:

PE = mgh

where g is the acceleration due to gravity.

Statement (v)

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

Correct Answer

Rate

Explanation: Power is the rate at which work is done or energy is transferred.

P = W / t

where W is the work done and t is the time taken.

“`

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.

✓ Answer: (iii) and (iv)

(i) Incorrect: Even at the highest point, the force of gravity acts downward on the ball. Therefore, the force is not zero.

(ii) Incorrect: The acceleration due to gravity continues to act downward. Thus, the acceleration is g ≈ 9.8 m/s², not zero.

(iii) Correct: At the highest point, the velocity of the ball is zero. Therefore, KE = ½mv² = 0.

(iv) Correct: At the highest point, the height of the ball is maximum. Hence, its gravitational potential energy PE = mgh is maximum.

💡 Key Concept: At the highest point, velocity = 0, but acceleration ≠ 0. Gravity continues to act on the ball.

QUESTION 4

For each of the following situations, identify the energy transformation that takes place:

A truck moving uphill.
Unwinding of a watch spring.
Photosynthesis in green leaves.
Water flowing from a dam.
Burning of a matchstick.
Explosion of a fire cracker.
Speaking into a microphone.
A glowing electric bulb.
A solar panel.

✓ Answer

Truck moving uphill: Chemical energy of fuel → Kinetic energy → Gravitational potential energy.
Unwinding of a watch spring: Potential energy stored in the spring → Kinetic energy → Mechanical energy of the watch.
Photosynthesis in green leaves: Solar energy (light energy) → Chemical energy stored in food.
Water flowing from a dam: Gravitational potential energy → Kinetic energy → Mechanical energy.
Burning of a matchstick: Chemical energy → Heat energy + Light energy.
Explosion of a fire cracker: Chemical energy → Heat energy + Light energy + Sound energy + Kinetic energy.
Speaking into a microphone: Sound energy → Electrical energy.
Glowing electric bulb: Electrical energy → Light energy + Heat energy.
Solar panel: Solar/light energy → Electrical energy.

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:

Height of building, h = 72.5 m
Acceleration due to gravity, g = 10 m/s²
Mass of student, m = 50 kg

✓ Complete Answer

(i) Gain in potential energy when lifted by elevator

Gravitational potential energy is:

PE = mgh

Substituting the values:

PE = 50 × 10 × 72.5

PE = 36,250 J

Therefore, the gain in potential energy is 36,250 J.

(ii) Gain in potential energy when the student climbs the stairs

The student reaches the same height, 72.5 m. Therefore:

PE = mgh = 50 × 10 × 72.5

PE = 36,250 J

Thus, the gain in potential energy while climbing the stairs is also 36,250 J.

(iii) Conclusion

In both cases, the student reaches the same final height. Hence, the gain in potential energy is the same, irrespective of whether the student is lifted vertically or climbs the staircase.

Conclusion: Gravitational potential energy depends on the height of the object, not on the path taken to reach that height.

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.

✓ Complete Answer

Let the height of each floor be h.

Energy required to lift a mass through height H is:

E = mgh

Energy for the 10th floor

Height = 10h

E₁ = mg(10h) = 10mgh

Energy for the 20th floor

Height = 20h

E₂ = mg(20h) = 20mgh

Therefore:

E₂ = 2E₁

So, the energy required is twice that required for the 10th floor, or 100% more energy.

Power comparison

Let the time taken to reach the 10th floor be t. The time taken to reach the 20th floor is 2t.

P₁ = E₁/t

P₂ = E₂/2t = 2E₁/2t = E₁/t

Therefore, P₂ = P₁.

Final Answer: Energy required = 2 times (100% more).
Power required = Same as before.

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.

✓ Complete Answer

1. Factors determining the energy required

The energy required to raise the flag is equal to the gain in its gravitational potential energy:

E = mgh

Therefore, the energy depends on:

Mass (m) of the flag.
Acceleration due to gravity (g).
Height (h) through which the flag is raised.

2. Does speed affect the amount of work done?

No. If the flag is raised to the same height, the gain in gravitational potential energy is the same.

Therefore, raising the flag slowly or quickly does not change the amount of work done, assuming there are no energy losses.

3. What happens to power when speed is doubled?

Power is the rate of doing work:

P = W/t

If the speed of raising the flag is doubled, the same distance is covered in half the time. Therefore:

P ∝ 1/t

Hence, when the speed is doubled, the power requirement becomes twice.

Final Answer: Energy depends on mass, gravity and height. Work done is independent of the speed of raising the flag, but doubling the speed doubles the power requirement.

QUESTION 8

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.

✓ Complete Answer

Since the energy supplied by the fuel is converted entirely into the kinetic energy of the scooter and riders:

KE = ½Mv²

First day

Mass of scooter = 100 kg
Mass of man = 60 kg

Total mass:

M₁ = 100 + 60 = 160 kg

Therefore:

E₁ = ½ × 160 × v² = 80v²

Second day

Mass of scooter = 100 kg
Mass of man = 60 kg
Mass of son = 40 kg

Total mass:

M₂ = 100 + 60 + 40 = 200 kg

Therefore:

E₂ = ½ × 200 × v² = 100v²

Ratio of fuel used

Since fuel consumption is directly proportional to the energy required:

Fuel₁ : Fuel₂ = E₁ : E₂

= 80v² : 100v²

Fuel₁ : Fuel₂ = 4 : 5

Thus, the fuel used on the first day : second day is 4 : 5.

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.

✓ Answer

For the seesaw to remain balanced, the clockwise moment must be equal to the anticlockwise moment.

Force × Distance = Force × Distance

Let the weight of the child be W. Therefore, the weight of the adult is 2W.

Let the child sit at a distance d from the fulcrum. If the adult sits at a distance D, then for balance:

W × d = 2W × D

Cancelling W:

d = 2D

Therefore, the child must sit at twice the distance from the fulcrum compared with the adult.

Balanced Seesaw

👦 Child

← 2d →

─────────────────────────────

▲

Fulcrum

← d →

Adult 👨

Final Answer:
Since the adult weighs twice as much as the child, the adult must sit at half the distance from the fulcrum compared with the child. Thus, if the adult sits at distance d, the child should sit at distance 2d.

QUESTION 10

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.)

✓ Complete Answer

(i) Sign of work done by gravity

Work done is given by:

W = F × s × cos θ

During the upward motion, the displacement of the ball is upward, while the gravitational force acts downward. Therefore, the angle between force and displacement is 180°.

Upward motion → Work done by gravity is negative.

During the downward motion, both the displacement and gravitational force are in the downward direction. Therefore, the angle between them is 0°.

Downward motion → Work done by gravity is positive.

(ii) Work done by air resistance

Given:

Mass, m = 2 kg
Initial velocity, u = 20 m s⁻¹
Height reached, h = 19.4 m
Acceleration due to gravity, g = 10 m s⁻²

Step 1: Initial kinetic energy

KE_initial = ½mu²

= ½ × 2 × (20)²

KE_initial = 400 J

Step 2: Gain in gravitational potential energy

PE = mgh

= 2 × 10 × 19.4

PE = 388 J

Step 3: Apply the work-energy theorem

The total work done on the ball equals the change in its kinetic energy. At the maximum height, the velocity of the ball is zero.

KE_final = 0

Therefore, the change in kinetic energy is:

ΔKE = 0 − 400 = −400 J

The work done by gravity is:

W_gravity = −mgh = −388 J

Let the work done by air resistance be W_air.

By the work-energy theorem:

W_gravity + W_air = ΔKE

−388 + W_air = −400

W_air = −12 J

✓ Final Answer

(i) Work done by gravity is negative during upward motion and positive during downward motion.

(ii) Work done by air resistance = −12 J.

The negative sign indicates that air resistance acts opposite to the direction of motion and therefore removes energy from the ball.

QUESTION 11

A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in 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?

Given:

Mass of block, m = 10.0 kg
Initial kinetic energy, K_i = 180 J
Friction is negligible.
The variable force acts in the direction of motion.

✓ Complete Answer

(i) Speed of the block at 0 m

The kinetic energy of a body is given by:

K = ½mv²

At 0 m, the kinetic energy is 180 J.

180 = ½ × 10 × v²

180 = 5v²

v² = 36

v = 6 m/s

Therefore, the speed of the block at 0 m is 6 m/s.

(ii) Speed of the block at 4 m

The work done by a variable force is equal to the area under the force–displacement graph.

Work done from 0 m to 4 m

From the graph:

From 0–1 m: triangular region
From 1–3 m: rectangular region
From 3–4 m: triangular region

Work from 0–1 m:

W₁ = ½ × base × height

W₁ = ½ × 1 × 50

W₁ = 25 J

Work from 1–3 m:

W₂ = length × breadth

W₂ = 2 × 50

W₂ = 100 J

Work from 3–4 m:

W₃ = ½ × base × height

W₃ = ½ × 1 × 50

W₃ = 25 J

Total Work Done

W = W₁ + W₂ + W₃

W = 25 + 100 + 25

W = 150 J

Using the Work–Energy Theorem

According to the work–energy theorem:

Work done = Change in kinetic energy

Therefore:

W = K_f − K_i

150 = K_f − 180

K_f = 330 J

Finding the speed at 4 m

At 4 m:

K_f = ½mv²

330 = ½ × 10 × v²

330 = 5v²

v² = 66

v = √66 ≈ 8.12 m/s

Does the block have negative acceleration?

No. The applied force is always in the direction of the block’s motion. Therefore, the force is either positive or zero.

Since friction is negligible:

F = ma

Thus, the acceleration is positive wherever the force is positive and becomes zero where the force becomes zero. It never becomes negative.

✓ Final Answer

(i) Speed at 0 m = 6 m/s

(ii) Speed at 4 m = √66 ≈ 8.12 m/s

Negative acceleration: No, the block does not have negative acceleration in any portion of its motion.

QUESTION

The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6 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?

Given:

Maximum height on Earth, h_E = 8 m
Moon’s gravity, g_M = g_E/6
The ball is thrown with the same initial velocity on Earth and Moon.

✓ Complete Answer

Step 1: Use the equation for maximum height

When a ball is thrown vertically upward, its final velocity at the maximum height becomes zero.

v² = u² − 2gh

At the maximum height, v = 0. Therefore:

0 = u² − 2gh

Hence:

u² = 2gh

Step 2: For the Earth

The astronaut can throw the ball to a height of 8 m:

u² = 2g_Eh_E

Since h_E = 8 m:

u² = 2g_E × 8

Step 3: For the Moon

The same ball is thrown with the same initial velocity. Let its maximum height on the Moon be h_M.

Since the gravitational acceleration on the Moon is:

g_M = g_E/6

Therefore:

u² = 2g_Mh_M

Since the initial velocity u is the same:

2g_Eh_E = 2g_Mh_M

Cancelling 2:

g_Eh_E = g_Mh_M

Substitute g_M = g_E/6 and h_E = 8 m:

g_E × 8 = (g_E/6) × h_M

Cancelling g_E:

8 = h_M/6

h_M = 48 m

✓ Final Answer

The ball will rise to a height of 48 m above the lunar surface.

Reason: The Moon’s gravitational acceleration is only one-sixth of Earth’s. For the same initial velocity, the maximum height is inversely proportional to gravitational acceleration. Therefore, the height on the Moon is 6 times the height on Earth.

“`html

CLASS 9 SCIENCE • WORK, ENERGY AND SIMPLE MACHINES

Question 13

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.

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

Answer

(i) Motion between A and B

From the speed–time graph, the speed of the car remains constant at 35 m s⁻¹ between A and B.

Therefore: The car moves with uniform speed between A and B. This represents the driver’s reaction time before the brakes are applied.

(ii) Kinetic Energy at A

Given:

Mass of car, m = 1000 kg
Speed at A, v = 35 m s⁻¹

Kinetic Energy

KE = ½mv²

KE = ½ × 1000 × (35)²

= 500 × 1225

= 612500 J

Therefore, the kinetic energy of the car at A is 6.125 × 10⁵ J.

(iii) Work Done by the Brakes

Between B and C, the brakes bring the car to rest.

Initial kinetic energy at B: 612500 J

Final kinetic energy at C: 0 J

Work done = Final KE − Initial KE

W = 0 − 612500

W = −612500 J

The negative sign indicates that the braking force acts opposite to the direction of motion.

(iv) Transformation of Kinetic Energy

When the brakes are applied, the kinetic energy of the moving car is mainly transformed into thermal energy (heat) due to friction in the brakes, tyres and road.

A small part of the energy is also transformed into sound energy.

Kinetic Energy → Heat Energy + Sound Energy

Quick Revision

A → B: Uniform speed = 35 m s⁻¹

KE at A: 6.125 × 10⁵ J

Work by brakes: −6.125 × 10⁵ J

Energy transformation: Kinetic energy → mainly heat energy + some sound energy

“`“`html

CLASS 9 SCIENCE • WORK, ENERGY AND SIMPLE MACHINES

Question 15

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.

Calculate the velocity of the coconut just before it hits the sand.
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

(i) Velocity just before hitting the sand

Given:

Mass of coconut, m = 1.5 kg
Height, h = 10 m
Acceleration due to gravity, g = 10 m s⁻²
Initial velocity, u = 0

Using the equation of motion:

v² = u² + 2gh

v² = 0 + 2 × 10 × 10

v² = 200

v = √200 = 10√2 ≈ 14.14 m s⁻¹

Answer: The velocity of the coconut just before it hits the sand is 14.14 m s⁻¹ downward.

(ii) Depth of the depression

Just before hitting the sand, the coconut has kinetic energy.

Kinetic energy:

KE = ½mv²

KE = ½ × 1.5 × 200

KE = 150 J

This entire kinetic energy is used to overcome the resistive force of the sand and create the depression.

Work done by the resistive force is:

Work = Force × displacement

150 = 3000 × d

d = 150 / 3000

d = 0.05 m = 5 cm

Answer: The depth of the depression made by the coconut is 0.05 m or 5 cm.

Quick Revision

(i) Velocity: 14.14 m s⁻¹ downward

(ii) Depth of depression: 0.05 m = 5 cm

“`

Conservation of Mechanical Energy

Question 14

The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown below. 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.

Potential Energy–Displacement Graph

Fig. 7.39 — Potential energy vs displacement

Answer

Step 1: Find the total mechanical energy

At O, the velocity is zero, so the kinetic energy is zero. Therefore, the total mechanical energy is equal to the potential energy at O.

Total Energy = PE + KE = 30 + 0 = 30 J

Using conservation of mechanical energy:

PE + KE = Constant
PE + ½mv² = 30 J

At P

From the graph:PE_P = 20 J
KE_P = 30 − 20 = 10 J
½ × 0.5 × v_P² = 10
v_P² = 40
v_P = 6.32 m s⁻¹

At Q

From the graph:PE_Q = 30 J
KE_Q = 30 − 30 = 0 J
Therefore, v_Q = 0 m s⁻¹

The ball comes momentarily to rest at Q.

At R

From the graph:PE_R = 40 J
KE_R = 30 − 40 = −10 J
Kinetic energy cannot be negative. Therefore, the ball does not have enough mechanical energy to reach R.

✕ The ball cannot reach R.

Final Answer

P
6.32 m s⁻¹

Q
0 m s⁻¹

R
Cannot reach

“`html

EXTRA QUESTION • WORK, ENERGY AND SIMPLE MACHINES

Question 1: State the Law of Conservation of Energy

State the law of conservation of energy. Explain it with the help of a falling object and a simple pendulum. Also derive mathematically how mechanical energy remains conserved in these situations.

Answer

1. Law of Conservation of Energy

Energy can neither be created nor destroyed.
It can only be transformed from one form to another.

Therefore, the total energy of an isolated system always remains constant.

Total Energy = Constant

In mechanical systems, energy commonly changes between potential energy and kinetic energy.

Mechanical Energy = Kinetic Energy + Potential Energy

KE + PE = Constant

2. Conservation of Energy in a Falling Object

Consider an object of mass m held at a height h above the ground. Let it be released from rest. Ignore air resistance.

At the initial position:

Velocity, u = 0
Height = h

Step 1: Energy at the top

Since the object is initially at rest, its kinetic energy is zero.

KE = ½mu² = 0

PE = mgh

Therefore, the total mechanical energy at the top is:

Total Energy = mgh

Step 2: Object falls through a distance

Suppose the object has fallen through a distance x. Its height above the ground is now:

Height = h − x

Using the equation of motion:

v² = u² + 2gx

Since u = 0:

v² = 2gx

Step 3: Kinetic Energy at this position

The kinetic energy of the object is:

KE = ½mv²

KE = ½m(2gx)

KE = mgx

Step 4: Potential Energy at this position

The object is now at height (h − x). Therefore:

PE = mg(h − x)

PE = mgh − mgx

Step 5: Total Mechanical Energy

Adding kinetic energy and potential energy:

KE + PE

= mgx + (mgh − mgx)

= mgh

Thus, at every position during the fall:

KE + PE = mgh = Constant

As the object falls, potential energy decreases and kinetic energy increases by exactly the same amount. Hence, total mechanical energy remains constant.

3. Conservation of Energy in a Simple Pendulum

Consider a simple pendulum consisting of a bob of mass m attached to a light, inextensible string. The bob is displaced to one side and released from rest.

Let the maximum vertical height of the bob above its lowest position be h.

Position A — Extreme Position

At the extreme position, the bob is momentarily at rest. Therefore:

v = 0

KE = 0
PE = mgh

Hence:

Total Energy = mgh

Position B — Intermediate Position

Suppose the bob is at an intermediate position at height y above the lowest point.

If the bob has fallen through a vertical distance (h − y), then using:

v² = 2g(h − y)

Therefore its kinetic energy is:

KE = ½mv²

= ½m[2g(h − y)]

KE = mg(h − y)

Its potential energy is:

PE = mgy

Therefore, total mechanical energy is:

KE + PE

= mg(h − y) + mgy

= mgh − mgy + mgy

= mgh

Position C — Lowest Position

At the lowest position, the height is taken as zero. Therefore:

NCERT Class 10 Mathematics Solutions

This page will guide you to the textbook and solutions, as provided by the National Council of Educational Research and Training (NCERT).

NCERT Maths All Chapters

Quick Revision

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

For each of the following situations, identify the energy transformation that takes place:

✓ Answer

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.

✓ Complete Answer

(i) Gain in potential energy when lifted by elevator

(ii) Gain in potential energy when the student climbs the stairs

(iii) Conclusion

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.

✓ Complete Answer

Energy for the 10th floor

Energy for the 20th floor

Power comparison

✓ Complete Answer

1. Factors determining the energy required

2. Does speed affect the amount of work done?

3. What happens to power when speed is doubled?

✓ Complete Answer

First day

Second day

Ratio of fuel used

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 ball of mass 2 kg is thrown up with a velocity of 20 m s−1.

✓ Complete Answer

(i) Sign of work done by gravity

(ii) Work done by air resistance

Step 1: Initial kinetic energy

Step 2: Gain in gravitational potential energy

Step 3: Apply the work-energy theorem

Given:

✓ Complete Answer

(i) Speed of the block at 0 m

(ii) Speed of the block at 4 m

Work done from 0 m to 4 m

Using the Work–Energy Theorem

Finding the speed at 4 m

Does the block have negative acceleration?

Given:

✓ Complete Answer

Step 1: Use the equation for maximum height

Step 2: For the Earth

Step 3: For the Moon

Question 13

Answer

(i) Motion between A and B

(ii) Kinetic Energy at A

(iii) Work Done by the Brakes

(iv) Transformation of Kinetic Energy

Quick Revision

Question 15

Answer

(i) Velocity just before hitting the sand

(ii) Depth of the depression

Quick Revision

Question 1: State the Law of Conservation of Energy

Answer

1. Law of Conservation of Energy

2. Conservation of Energy in a Falling Object

Step 1: Energy at the top

Step 2: Object falls through a distance

Step 3: Kinetic Energy at this position

Step 4: Potential Energy at this position

Step 5: Total Mechanical Energy

3. Conservation of Energy in a Simple Pendulum

Position A — Extreme Position

Position B — Intermediate Position

Position C — Lowest Position

A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹.