Class 9 - Chapter 7: Work, Energy, and Simple Machines

Exercise Solution

1. State whether True or False

(i) False

(ii) True

(iii) True

(iv) False

(v) True

2. Fill in the blanks

(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 is

KE = ½mv²

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

PE = mgh

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

3. Ball thrown upwards

Correct statements:

(iii) Its kinetic energy is zero. ✓

(iv) Its potential energy is maximum. ✓

At the highest point, velocity becomes zero, hence kinetic energy becomes zero and potential energy becomes maximum.

4. Energy transformations

5. Student climbing a building

Given:

h = 72.5 m

m = 50 kg

g = 10 m s^-2

(i) Lifted by elevator

PE = mgh

= 50 × 10 × 72.5

= 36250 J

(ii) Climbing stairs

PE = mgh

= 50 × 10 × 72.5

= 36250 J

(iii) Conclusion

Potential energy depends only on vertical height gained, not on the path followed.

6. Crane lifting a mass

Potential energy:

PE = mgh

When height doubles from 10th floor to 20th floor:

Energy required becomes double.

Time also doubles.

Power = Energy / Time

Power remains the same.

Answer:

Energy required = 2 times

Power required = same

7. Raising a flag

Work done:

W = mgh

Therefore work depends on:

• Mass of flag
• Height of pole
• Acceleration due to gravity

Raising slowly or quickly does not change work done.

If speed doubles, time becomes half.

Hence power required doubles.

8. Fuel consumption ratio

Day 1:

Total mass = 60 + 100 = 160 kg

Day 2:

Total mass = 60 + 100 + 40 = 200 kg

Kinetic Energy:

KE = ½mv²

Since velocity is same,

Fuel ∝ Mass

Ratio:

160 : 200

= 4 : 5

Answer: Fuel consumed ratio = 4 : 5

9. Seesaw problem

Let weight of child = W

Weight of adult = 2W

For balance:

Clockwise moment = Anticlockwise moment

W × d_child = 2W × d_adult

d_child = 2d_adult

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

10. Ball thrown upward

m = 2 kg

u = 20 m s^-1

(i) Sign of work done by gravity

Upward motion → Negative work

Downward motion → Positive work

(ii) Work done by air resistance

Maximum height without air resistance:

H = u² / 2g

= 20² / (2 × 10)

= 20 m

Actual height = 19.4 m

Loss in PE:

= mg(20 − 19.4)

= 2 × 10 × 0.6

= 12 J

Answer: Work done by air resistance = −12 J

11. Block on frictionless floor

From Fig. 7.37:

Work done = Area under Force–Displacement graph

Triangle (0–1 m):

= ½ × 1 × 50

= 25 J

Rectangle (1–3 m):

= 2 × 50

= 100 J

Triangle (3–4 m):

= ½ × 1 × 50

= 25 J

Total work:

= 25 + 100 + 25

= 150 J

Given KE at 0 m = 180 J

Using Work-Energy theorem:

KE at 4 m = 180 + 150

= 330 J

Speed at 0 m:

180 = ½ × 10 × v²

v = 6 m s^-1

Speed at 4 m:

330 = ½ × 10 × v²

v = √66

≈ 8.12 m s^-1

Acceleration is never negative because force is always positive.

12. Ball thrown on the Moon

g_moon = g/6

Maximum height:

H ∝ 1/g

H_moon = 6 × H_earth

= 6 × 8

= 48 m

Answer: Height reached on Moon = 48 m

13. Car stopping

(i)

Between A and B the car moves with constant speed.

(ii)

At A:

Mass = 1000 kg

Speed = 35 m s^-1

KE = ½mv²

= ½ × 1000 × 35²

= 612500 J

(iii)

Work done by brakes:

= Change in KE

= 0 − 612500

= −612500 J

(iv)

Kinetic energy is transformed into heat energy due to friction in brakes.

14. Potential Energy Graph

At O:

KE = 0

PE = 30 J

Total Energy = 30 J

From graph:

PE at P = 10 J

PE at Q = 40 J

At P:

KE = 30 − 10

= 20 J

v = √(2KE/m)

= √(40/0.5)

= √80

≈ 8.94 m s^-1

At Q:

PE = 40 J > Total Energy

Hence the ball cannot reach Q.

15. Coconut falling from a tree

m = 1.5 kg

h = 10 m

g = 10 m s^-2

(i) Velocity before hitting sand

v² = 2gh

= 2 × 10 × 10

= 200

v = √200

≈ 14.14 m s^-1

(ii) Depth of depression

Potential energy:

PE = mgh

= 1.5 × 10 × 10

= 150 J

Work done against sand:

F × d = 150

3000 × d = 150

d = 150/3000

d = 0.05 m

= 5 cm

Answer: Depth of depression = 5 cm