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The force constant of a weightless sprin...

The force constant of a weightless spring is `16 N m^(-1)`. A body of mass `1.0 kg` suspended from it is pulled down through `5 cm` and then released. The maximum energy of the system (spring + body) will be

A

`2 xx 10^(-2)`J

B

`4 xx 10^(-2)`J

C

`8 xx 10^(-2)`J

D

`16 xx 10^(-2)`J

Text Solution

Verified by Experts

The correct Answer is:
A
(a)`omega=sqrt((k)/(m))=sqrt((16)/(1))=4 "rad" s^(-1)`
Now , `K_("max")=(1)/(2)momega^(2)A^(2)`
`=(1)/(2)xx1xx(4)^(2)(5xx 10^(-2))^(2)`
`=2 xx 10^(-2)`J
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