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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(KE) = max(PE)
`=1/2 kx^2 =1/2 xx 16 xx(5 xx10^(-2))^(2) = 2 xx10^(-2) J`
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