Oscillatory Motion :: Discussion


1.A spring of force constant 800N/m has an extension of 5cm. The work done in extending it from 5cm to 15cm is?
A.8J
B.16J
C.24J
D.32J
Answer:  Option  A
Explanation:

At x1 = 5 cm,
U1=1/2×k(x1)2=1/2×800×0.052=1J
At x2=15cm,
U2=1/2×k(x2)2=1/2×800×0.152=9J
W=U2-U1=9-1=8J.

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Published by:Michael Daani

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