Physics Elasticity Question 65
Question: K is the force constant of a spring. The work done in increasing its extension from $ l _{1} $ to $ l _{2} $ will be
[MP PET 1995; MP PMT 1996]
Options:
A) $ K(l _{2}-l _{1}) $
B) $ \frac{K}{2}(l _{2}+l _{1}) $
C) $ K(l _{2}^{2}-l _{1}^{2}) $
D) $ \frac{K}{2}(l _{2}^{2}-l _{1}^{2}) $
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Answer:
Correct Answer: D
Solution:
At extension $ l _{1} $ , the stored energy $ =\frac{1}{2}Kl _{1}^{2} $ At extension $ l _{2} $ , the stored energy$ =\frac{1}{2}Kl _{2}^{2} $ Work done in increasing its extension from $ l _{1} $ to $ l _{2} $
$ =\frac{1}{2}K(l _{2}^{2}-l _{1}^{2}) $
BETA
