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When a long spring is stretched by 2cm, ...

When a long spring is stretched by 2cm, its potential energy is U. If the spring is stretched by 10cm, the potential energy stored in it will be

A

25U

B

`U/25`

C

`5U`

D

`U/5`

Text Solution

Verified by Experts

The correct Answer is:
A
Loss in K.E. of mass=Gain in Elastic potential energy of the spring `therefore 1/2MV^(2)=1/2Kx^(2)`
`therefore 0.5 xx 1.5 xx 1.5 =50 xx x^(2)`
`x^(2)=3/2 xx 3/2 xx 1/2 xxx 1/50=9/(400)`
`x=(3)/(20)=(15)/(100) =0.15m`
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