First slide

Potential energy

Question

A spring of force constant 10 N/m has initial stretch 0.2 m. In changing the stretch to 0.25 m, the increase of PE is about :

Easy

A

0. 1 joule
B

0.2 joule
C

0.3 joule
D

0.5 joule

Solution

Increase in PE = $= \frac{1}{2} {kx}_{2}^{2} - {kx}_{1}^{2} = \frac{1}{2} (10) [0 {.25}^{2} - 0 {.2}^{2}] = 0 .1125 J$

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