Vidaara.orgClass 11 · Physics
CodeVID-P11-05-PEC-01
Potential Energy & Conservation — 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.
Gravitational potential energy near the Earth is given by:
- A.$\frac{1}{2}mv^2$
- B.$mgh$
- C.$\frac{1}{2}kx^2$
- D.$mv$
2.
The work done by a conservative force over a closed path is:
- A.maximum
- B.zero
- C.negative
- D.infinite
3.
Spring potential energy depends on extension as:
- A.$x$
- B.$x^2$
- C.$\sqrt{x}$
- D.$\frac{1}{x}$
4.
Which force is conservative?
- A.friction
- B.air resistance
- C.spring force
- D.viscous drag
5.
For a freely falling body, the sum $KE+PE$:
- A.increases
- B.decreases
- C.stays constant
- D.is zero
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Find the PE of a 5 kg body at a height of 10 m ($g=10\ \text{m/s}^2$).
7.
Define a conservative force.
8.
A spring of $k=100\ \text{N/m}$ is compressed by 0.2 m. Find the stored energy.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A stone is dropped from a height of 45 m. Using energy conservation, find its speed on reaching the ground ($g=10\ \text{m/s}^2$).
10.
Show that the elastic PE of a spring is $\frac{1}{2}kx^2$ using the area under its force-extension graph.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State and prove the conservation of mechanical energy for a body of mass $m$ falling freely from height $H$, showing that $KE+PE$ is the same at the top, at a point distance $x$ below the top, and at the ground.
Answer Key
Section A — Multiple Choice Questions
- (B) $mgh$
- (B) zero
- (B) $x^2$
- (C) spring force
- (C) stays constant
Section B — Short Answer (2 marks)
- $U=mgh=5\times 10\times 10=500\ \text{J}$.
- A force whose work depends only on the end points, not the path, and is zero over a closed loop.
- $U=\frac{1}{2}\times 100\times (0.2)^2=2\ \text{J}$.
Section C — Short Answer (3 marks)
- $v=\sqrt{2gh}=\sqrt{2\times 10\times 45}=\sqrt{900}=30\ \text{m/s}$.
- The graph of $F=kx$ is a straight line; the area under it up to $x$ is a triangle of base $x$ and height $kx$, so $U=\frac{1}{2}(x)(kx)=\frac{1}{2}kx^2$.
Section D — Long Answer (5 marks)
- At the top: $E=mgH$. After falling $x$: $v^2=2gx$, so $KE=mgx$, $PE=mg(H-x)$, sum $=mgH$. At the ground: $v^2=2gH$, $KE=mgH$, $PE=0$, sum $=mgH$. Hence $KE+PE=mgH$ throughout.
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