First slide

Work

Question

A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about

Moderate

A

0.1 joule
B

0.2 joule
C

0.3 joule
D

0.5 joule

Solution

$Δ P . E . = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 10 [{(0 .25)}^{2} - {(0 .20)}^{2}]$
$= 5 \times 0 .45 \times 0 .05 = 0 .1 J$

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