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CodeVID-P9-04-CT
Work and Energy — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Take $g=10\,\text{m/s}^2$ unless stated otherwise. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The SI unit of work is the
- A.newton
- B.joule
- C.watt
- D.pascal
2.
Kinetic energy is given by
- A.$mgh$
- B.$\frac{1}{2}mv^2$
- C.$Fs$
- D.$\frac{W}{t}$
3.
Gravitational potential energy is given by
- A.$\frac{1}{2}mv^2$
- B.$mgh$
- C.$Fs\cos\theta$
- D.$\frac{W}{t}$
4.
The SI unit of power is the
- A.joule
- B.watt
- C.newton
- D.kWh
5.
$1\,\text{kWh}$ equals
- A.$3600\,\text{J}$
- B.$3.6\times10^6\,\text{J}$
- C.$1000\,\text{J}$
- D.$1\,\text{J}$
6.
Work done by a force perpendicular to displacement is
- A.maximum
- B.zero
- C.negative
- D.1 J
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
A force of 12 N moves a body 5 m along its direction. Find the work done.
8.
Find the kinetic energy of a 3 kg body moving at 6 m/s.
9.
A motor does 1500 J of work in 5 s. Find its power.
10.
State the law of conservation of energy.
Section C — Short Answer (3 marks)
3 × 3 = 9 marks
11.
A 50 kg load is lifted 4 m. Find the work done and the potential energy gained.
12.
A 2 kg ball is dropped from 5 m. Find its speed at the ground using energy conservation.
13.
A 1 kW geyser runs for 3 hours. Find the energy used in kWh and joules.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Explain positive, negative and zero work with one example each, and define one joule.
15.
A 1 kg ball is dropped from 20 m. Find PE and KE at the top, at 10 m and at the ground, and show energy is conserved. Take $g=10\,\text{m/s}^2$.
Answer Key
Section A — Multiple Choice Questions
- (B) joule
- (B) $\frac{1}{2}mv^2$
- (B) $mgh$
- (B) watt
- (B) $3.6\times10^6\,\text{J}$
- (B) zero
Section B — Short Answer (2 marks)
- $W=Fs=12\times5=60\,\text{J}$.
- $E_k=\frac{1}{2}\times3\times6^2=54\,\text{J}$.
- $P=\frac{W}{t}=\frac{1500}{5}=300\,\text{W}$.
- Energy can neither be created nor destroyed; it can only be transformed from one form to another, keeping the total constant.
Section C — Short Answer (3 marks)
- $W=mgh=50\times10\times4=2000\,\text{J}$; PE gained $=2000\,\text{J}$.
- $v^2=2gh=2\times10\times5=100\Rightarrow v=10\,\text{m/s}$.
- Energy $=1\times3=3\,\text{kWh}=3\times3.6\times10^6=1.08\times10^7\,\text{J}$.
Section D — Long Answer (5 marks)
- Positive work: force along motion, e.g. pushing a cart ($\theta<90^{\circ}$). Negative work: force opposing motion, e.g. friction ($\theta=180^{\circ}$). Zero work: no displacement, or force perpendicular to motion ($\theta=90^{\circ}$). One joule is the work done when a 1 N force moves a body 1 m in its direction.
- Top: PE $=200\,\text{J}$, KE $=0$, total 200 J. At 10 m: PE $=100\,\text{J}$, KE $=100\,\text{J}$, total 200 J. Ground: PE $=0$, KE $=200\,\text{J}$, total 200 J. Total mechanical energy stays 200 J, so energy is conserved.
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