Vidaara.orgClass 11 · Physics
CodeVID-P11-05-CH-01
Work, Energy and Power — Full Chapter 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
6 × 1 = 6 marks
1.
Work is a:
- A.vector
- B.scalar
- C.tensor
- D.unit vector
2.
The work done by friction is generally:
- A.positive
- B.negative
- C.zero
- D.infinite
3.
Kinetic energy depends on:
- A.position
- B.speed
- C.height
- D.shape
4.
Spring PE is proportional to:
- A.$x$
- B.$x^2$
- C.$\sqrt{x}$
- D.$\frac{1}{x}$
5.
The unit of power is the:
- A.joule
- B.watt
- C.newton
- D.ohm
6.
In an elastic collision, $e$ equals:
- A.$0$
- B.$0.5$
- C.$1$
- D.$2$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
A force of 25 N moves a body 4 m along its direction. Find the work done.
8.
Find the KE of a 3 kg body moving at 6 m/s.
9.
A spring ($k=150\ \text{N/m}$) is stretched 0.2 m. Find the stored PE.
10.
A machine does 8000 J of work in 16 s. Find its power.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
A variable force $F=(4x+2)\ \text{N}$ acts from $x=0$ to $x=2\ \text{m}$. Find the work done.
12.
A 5 kg body at 6 m/s collides with and sticks to a 1 kg body at rest. Find the common velocity and the KE lost.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
State and derive the work-energy theorem for a constant force, then apply it to find the stopping distance of a 1200 kg car moving at 20 m/s under a retarding force of 6000 N.
14.
Prove the conservation of mechanical energy for a freely falling body and verify it at the top, midpoint and bottom of a 40 m fall ($g=10\ \text{m/s}^2$, mass $m$).
Answer Key
Section A — Multiple Choice Questions
- (B) scalar
- (B) negative
- (B) speed
- (B) $x^2$
- (B) watt
- (C) $1$
Section B — Short Answer (2 marks)
- $W=25\times 4=100\ \text{J}$.
- $KE=\frac{1}{2}\times 3\times 6^2=54\ \text{J}$.
- $U=\frac{1}{2}\times 150\times (0.2)^2=3\ \text{J}$.
- $P=\frac{8000}{16}=500\ \text{W}$.
Section C — Short Answer (3 marks)
- $W=\int_0^2 (4x+2)\,dx=[2x^2+2x]_0^2=8+4=12\ \text{J}$.
- $v=\frac{5\times 6}{6}=5\ \text{m/s}$; initial KE $=90\ \text{J}$, final KE $=\frac{1}{2}\times 6\times 5^2=75\ \text{J}$, KE lost $=15\ \text{J}$.
Section D — Long Answer (5 marks)
- $W_{net}=\Delta KE$; derivation from $v^2=u^2+2as$ gives $Fs=\frac{1}{2}mv^2-\frac{1}{2}mu^2$. For braking, $6000\times s=\frac{1}{2}\times 1200\times 20^2=240000\ \text{J}$, so $s=\frac{240000}{6000}=40\ \text{m}$.
- Top: $E=mg(40)=400m$. Midpoint (20 m, $v^2=2g\times20=400$): $KE=200m$, $PE=200m$, sum $=400m$. Bottom ($v^2=2g\times40=800$): $KE=400m$, $PE=0$, sum $=400m$. Mechanical energy is constant at $400m\ \text{J}$.
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