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CBSE · Class 9 · Science

Work, Energy, and Simple Machines

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A force of 35 N moves an object 22 m in the direction of the force. The work done is:
  1. 1.59 J
  2. 57 J
  3. 1,540 J
  4. 770 J
2.
The kinetic energy of a 51 kg body moving at 20 m/s is:
  1. 10,200 J
  2. 20,400 J
  3. 510 J
  4. 1,020 J
3.
The potential energy of a 16 kg object at a height of 2 m is (g = 9.8 m/s2):
  1. 156.8 J
  2. 32 J
  3. 176.4 J
  4. 313.6 J
4.
3,424 J of work is done in 8 s. The power is:
  1. 3424 W
  2. 27,392 W
  3. 428 W
  4. 436 W
5.
The SI unit of work is:
  1. pascal (Pa)
  2. watt (W)
  3. joule (J)
  4. newton (N)
6.
10 kWh of energy in joules is:
  1. 3.6 × 107 J
  2. 3.6 × 106 J
  3. 3.6 × 104 J
  4. 10,000 J
7.
A 2000 W appliance is used for 2 hours a day for 20 days. The energy used is:
  1. 800 kWh
  2. 80,000 kWh
  3. 4 kWh
  4. 80 kWh
8.
A ramp is 9 m long and 1 m high. Ignoring friction, the effort needed to push a 3,600 N load up it is:
  1. 401 N
  2. 3,600 N
  3. 400 N
  4. 800 N
Section B — Short answer8 × 2 = 16 marks
9.
Find the work done in lifting a 48 kg load through a height of 16 m. Take g = 9.8 m/s2.
10.
A 29 kg body speeds up from 2 m/s to 10 m/s. Find the work done on it. (Give the answer to 2 decimal places where needed.)
11.
If its mass is multiplied by 4 and the speed of a body is divided by 5, what happens to its kinetic energy?
12.
A motor lifts 180 kg through 21 m in 30 s. Find its power. Take g = 9.8 m/s2. (Give the answer to 2 decimal places where needed.)
13.
A house uses 301 units of electricity in a month. Find the bill at Rs 6.5 per unit.
14.
A porter carries a 38 kg load on the head and walks 142 m along a level road. How much work is done against gravity?
15.
A ball falls from rest through a height of 122.5 m. Find its speed just before it hits the ground. Take g = 9.8 m/s2.
16.
Using a lemon squeezer (a lever of the second class), a load of 120 N is 50 cm from the fulcrum and the effort is applied 400 cm from it. Find the effort needed and the mechanical advantage.
Section C — Word problems4 × 4 = 16 marks
17.
A family uses 6 bulbs of 60 W for 4 hours a day and a 75 W fan for 6 hours a day. Find the electricity used in 30 days and the cost at Rs 5 per unit.
18.
A car of mass 1,700 kg speeds up from 18 km/h to 90 km/h. Find the work done by its engine (ignore friction).
19.
A crane lifts a load of 1,700 kg through 14 m in 40 s. Find the work done and the power of the crane. Take g = 9.8 m/s2. (Give the answer to 2 decimal places where needed.)
20.
A worker pushes a box of weight 1,440 N up a ramp 3 m long onto a platform 1 m high, using an effort of 600 N along the ramp. Find the useful work done, the work put in, the efficiency of the ramp and its mechanical advantage.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 9 · Science

Answer key

Work, Energy, and Simple Machines · AW-9S-WEN-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 770 JW = F × s = 35 × 22
2(a) 10,200 J½mv2 = ½ × 51 × 400
3(d) 313.6 Jmgh = 16 × 9.8 × 2
4(c) 428 WP = W ÷ t = 3424 ÷ 8
5(c) joule (J)—
6(a) 3.6 × 107 J1 kWh = 3.6 × 106 J; × 10
7(d) 80 kWh2000 × 2 × 20 ÷ 1000
8(c) 400 Neffort = load × height ÷ length = 3600 × 1 ÷ 9
97,526.4 JW = mgh = 48 × 9.8 × 16
101,392 Jwork = change in KE = ½ × 29 × (100 − 4)
11it becomes 4/25 times what it wasKE ∝ mv2: 4 × (1/5)2 = 4/25
121,234.8 WP = mgh ÷ t = 180 × 9.8 × 21 ÷ 30
13Rs 1,956.5301 × 6.5
140 Jgravity acts downwards and the motion is horizontal, at 90° to the force
1549 m/smgh = ½mv2, so v = √(2gh) = √(2 × 9.8 × 122.5) = √2401
1615 N; MA = 8effort × 400 = 120 × 50, so effort = 15; MA = load ÷ effort = 120 ÷ 15
1756.7 kWh; Rs 283.5(6 × 60 × 4 + 75 × 6) × 30 ÷ 1000 kWh
18510,000 J18 km/h = 5 m/s, 90 km/h = 25 m/s; work = ½ × 1700 × (625 − 25)
19233,240 J; 5,831 WW = mgh; P = W ÷ 40
20useful work 1,440 J; work put in 1,800 J; efficiency 80%; MA = 2.4useful work = 1440 × 1; work put in = 600 × 3; efficiency = 1440 ÷ 1800 × 100; MA = load ÷ effort = 1440 ÷ 600