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The potential energy of a particle is gi...

The potential energy of a particle is given by formula `U=100-5x + 100x^(2), where `U` and are .

A

At `0.05m` form the origin is 50 ms^(2).

B

At 0.05 m from the mean position is 100 ms^(2).

C

At 0.05 m feom the origin is 150 ms^(2).

D

At `0.05 m` from the mean position is `200 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`F =-(dU)/(dx) =5-200x`
At origin, `x=0`
`F=5N`
`a=F/m =5/(0.1) =50 m//s^(2)`
Mean position is `F=0`
or at, `x=5/(200) =0.025 m`
`a=F/m =(5-200x)/(0.1) =(50-2000x)`
At `0.05 m` from the origin,
`x=+0.05 m`
`x = - 0.05 m`
Substituting in Eq. (i), we have,
`|a|=150m//s^(2)`
or `=50 m//s^(2)`
At `0.05 m` from the mean position means,
`x=0.075 `
or `x=-0.025 m`
Substituting in Eq. (i) we have,
`|a|=100 m//s^(2)`.
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