5.Work, Energy, Power and Collision
easy

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.....$joule$

A

$0.1$

B

$0.2$

C

$0.3$

D

$0.5$

Solution

(a)$\Delta$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$

Standard 11
Physics

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