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Work, Energy and Power
Potential Energy of a Spring
7 questions from 7 papers, each with its answer.
- questions
- 7
- papers
- 7
- years asked, 2017–2026
- 1 of 10
- of Work, Energy and Power
- 9.86%
Appears in
Concepts tested
- Conservation of energy with a spring
- Conservation of mechanical energy
- Elastic potential energy
- elastic potential energy ∝ x²
- Energy stored for equal force
- Spring potential energy proportional to x squared
- Spring work: same extension vs same force
All Questions
2023 1 question
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The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm, potential energy stored in it will be:
Spring PE is $\frac{1}{2}kx^2$, so 4 times the stretch gives $16U$.
Potential energy stored in the spring $=\frac{1}{2}kx^2$ Now $\frac{1}{2}k(2)^2=U$ and $\frac{1}{2}k(8)^2=U'$ (say) $\Rightarrow U'=\frac{64}{4}U=16U$
2015 AIPMT-I 1 question
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Two similar springs P and Q have spring constants $K_P$ and $K_Q$, such that $K_P > K_Q$. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs $W_P$ and $W_Q$ are related as in case (a) and case (b), respectively:Given $K_P > K_Q$.
Case (a), same extension x: $W = \frac{1}{2}Kx^2 \Rightarrow W \propto K$, so $W_P > W_Q$.
Case (b), same force F: $x = \frac{F}{K}$, so $W = \frac{F^2}{2K} \Rightarrow W \propto \frac{1}{K}$, so $W_Q > W_P$.
2009 AIPMT 1 question
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A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will beThe block is released from rest and comes momentarily to rest again at the maximum extension x, so its kinetic energy is zero at both positions.
Loss of gravitational potential energy = gain in elastic potential energy of the spring.
$Mgx = \frac{1}{2}kx^2$
$x = \frac{2Mg}{k}$
(This is twice the equilibrium extension Mg/k, because the block overshoots the equilibrium position and oscillates about it.)
2006 AIPMT-PRE 1 question
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The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm the potential energy stored in it is :-Potential energy stored in a spring stretched by $x$: $U = \dfrac{1}{2}Kx^2$
$U = \dfrac{1}{2}K(2)^2$ and $U' = \dfrac{1}{2}K(8)^2$
$\Rightarrow \dfrac{U'}{U} = \left(\dfrac{8}{2}\right)^2 = 16$
$\Rightarrow U' = 16U$
2004 AIPMT 1 question
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A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be :-
The surface is smooth, so mechanical energy is conserved. At maximum compression the mass momentarily stops, and all its kinetic energy is stored as elastic potential energy of the spring.
According to the question: $\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2$
$\Rightarrow x = v\sqrt{\dfrac{m}{k}}$
$x = 1.5\sqrt{\dfrac{0.5}{50}} = 1.5\sqrt{0.01} = 1.5 \times 0.1$
$x = 0.15$ m
So the maximum compression of the spring is 0.15 m.
2003 AIPMT 1 question
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When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be :Potential energy stored in a stretched spring: $U = \dfrac{1}{2}kx^2$, so $U \propto x^2$.
$U = \dfrac{1}{2}k(2)^2$ and $U' = \dfrac{1}{2}k(10)^2$
$\dfrac{U'}{U} = \dfrac{10^2}{2^2} = 25$
$U' = 25U$
2001 AIPMT 1 question
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Two springs A and B having spring constants $K_A$ and $K_B$ ($K_A = 2K_B$) are stretched by applying force of equal magnitude. If energy stored in spring A is E then energy stored in B will be : -Energy stored in a spring: $E = \dfrac{1}{2}kx^2$. With force $F = kx$, $x = \dfrac{F}{k}$:
$E = \dfrac{kx^2}{2} = \dfrac{k^2x^2}{2k} = \dfrac{F^2}{2k}$
The force is equal, so $E \propto \dfrac{1}{k}$.
$\dfrac{E_B}{E_A} = \dfrac{K_A}{K_B} = 2 \Rightarrow E_B = 2E$