Vidaara.orgClass 11 · Physics
CodeVID-P11-05-WET-01
Work & the Work-Energy Theorem — Assignment
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.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The SI unit of work is the:
- A.newton
- B.joule
- C.watt
- D.pascal
2.
Work done is negative when $\theta$ is:
- A.$0^\circ$
- B.$45^\circ$
- C.$90^\circ$
- D.$180^\circ$
3.
Kinetic energy in terms of momentum is:
- A.$\frac{p}{2m}$
- B.$\frac{p^2}{2m}$
- C.$\frac{p^2}{m}$
- D.$2mp^2$
4.
The work done by a centripetal force in uniform circular motion is:
- A.positive
- B.negative
- C.zero
- D.infinite
5.
Work done equals change in:
- A.momentum
- B.kinetic energy
- C.speed
- D.acceleration
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
A force of 20 N moves a body 5 m along its direction. Find the work done.
7.
State the work-energy theorem.
8.
Find the kinetic energy of a 4 kg body moving at 10 m/s.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A variable force $F=(2x+1)\ \text{N}$ acts on a body. Find the work done from $x=0$ to $x=3\ \text{m}$.
10.
A 5 kg body moving at 4 m/s is brought to rest. Using the work-energy theorem, find the work done by the stopping force.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the work-energy theorem for a constant force and use it to explain why a heavier vehicle needs a longer braking distance for the same speed.
Answer Key
Section A — Multiple Choice Questions
- (B) joule
- (D) $180^\circ$
- (B) $\frac{p^2}{2m}$
- (C) zero
- (B) kinetic energy
Section B — Short Answer (2 marks)
- $W=20\times 5\times\cos 0^\circ=100\ \text{J}$.
- The net work done by all forces on a body equals the change in its kinetic energy: $W_{net}=\Delta KE$.
- $KE=\frac{1}{2}\times 4\times 10^2=200\ \text{J}$.
Section C — Short Answer (3 marks)
- $W=\int_0^3 (2x+1)\,dx=[x^2+x]_0^3=9+3=12\ \text{J}$.
- $W=\Delta KE=0-\frac{1}{2}\times 5\times 4^2=-40\ \text{J}$.
Section D — Long Answer (5 marks)
- From $v^2=u^2+2as$, $\frac{1}{2}mv^2-\frac{1}{2}mu^2=mas=Fs=W_{net}$. For braking, $|W|=\frac{1}{2}mu^2=Fs$, so $s=\frac{mu^2}{2F}$; greater mass means greater stopping distance for the same retarding force and speed.
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