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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

`2xx10^(-2)J`

B

`4xx10^(-2)J`

C

`8xx10^(-2)J`

D

`16xx10^(-2)J`

Text Solution

Verified by Experts

The correct Answer is:
A
Max. K.E. of the system = Max. P.E. of the system
`(1)/(2)kx^(2)=(1)/(2)xx(16)xx(5xx10^(-2))^(2)=2xx10^(-2)J`
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