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When a spring a stretched through a dist...

When a spring a stretched through a distance x, it exerts a force given by `F=(-5x-16x^(3))N`. What is the work done, when the spring is stretched from 0.1 m to 0.2 m?

A

`8.7xx10^(-2)J`

B

`12.2xx10^(-2)J`

C

`8.7xx10^(-1)J`

D

`12.2xx10^(-1)J`

Text Solution

Verified by Experts

The correct Answer is:
A
`F=-5x-16x^(3)=-(5+16x^(2))x=-kx`
`therefore k=5+16x^(2)`
Work done, `W=(1)/(2)k_(2)x_(2)^(2)-(1)/(2)k_(1)x_(1)^(2)`
`=(1)/(2)[5+16(0.2)^(2)](0.2)^(2)-(1)/(2)[5+16(0.1)^(2)](0.1)^(2)`
`=2.82xx4xx10^(-2)-2.58xx10^(-2)=8.7xx10^(-2)J`.
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