Vidaara.orgClass 9 · Physics
CodeVID-P9-04-WRK-01
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- 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
5 × 1 = 5 marks
1.
Work is a
- A.vector quantity
- B.scalar quantity
- C.force
- D.unit
2.
$1\,\text{J}$ equals
- A.$1\,\text{N/m}$
- B.$1\,\text{N}\times1\,\text{m}$
- C.$1\,\text{kg/m}$
- D.$1\,\text{N/s}$
3.
If displacement is zero, the work done is
- A.maximum
- B.negative
- C.zero
- D.1 J
4.
Work done by a force opposing motion is
- A.positive
- B.negative
- C.zero
- D.undefined
5.
$W=Fs\cos\theta$ gives zero work when $\theta=$
- A.$0^{\circ}$
- B.$45^{\circ}$
- C.$90^{\circ}$
- D.$180^{\circ}$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Define 1 joule of work.
7.
A force of 10 N moves a body 3 m along its direction. Find the work done.
8.
State the two conditions necessary for work to be done.
9.
Why is the work done by Earth's gravity on a satellite in circular orbit zero?
Section C — Short Answer (3 marks)
3 × 3 = 9 marks
10.
A box is lifted 2 m by a force equal to its weight of 60 N. Find the work done and state its sign.
11.
A 40 N friction force acts on a body that slides 5 m. Find the work done by friction.
12.
A 20 N force at $60^{\circ}$ to the horizontal drags an object 3 m. Find the work done ($\cos60^{\circ}=0.5$).
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
13.
Explain positive, negative and zero work with one everyday example of each, and write the formula $W=Fs\cos\theta$ explaining each symbol.
Answer Key
Section A — Multiple Choice Questions
- (B) scalar quantity
- (B) $1\,\text{N}\times1\,\text{m}$
- (C) zero
- (B) negative
- (C) $90^{\circ}$
Section B — Short Answer (2 marks)
- Work done when a force of 1 N moves a body 1 m in the direction of the force.
- $W=Fs=10\times3=30\,\text{J}$.
- A force must act on the body, and the body must be displaced.
- Gravity is perpendicular to the displacement, so $W=Fs\cos90^{\circ}=0$.
Section C — Short Answer (3 marks)
- $W=Fs=60\times2=120\,\text{J}$; positive work (force and motion both upward).
- $W=-Fs=-40\times5=-200\,\text{J}$ (negative work).
- $W=Fs\cos\theta=20\times3\times0.5=30\,\text{J}$.
Section D — Long Answer (5 marks)
- Positive: pushing a box forward ($\theta<90^{\circ}$). Negative: friction on a moving trolley ($\theta=180^{\circ}$). Zero: carrying a load on level ground ($\theta=90^{\circ}$) or no displacement. In $W=Fs\cos\theta$, $F$ is force (N), $s$ displacement (m), $\theta$ angle between them; $W$ in joules.
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