The potential energy U of a particle of mass m=1kg moving in x−y plane is given by U=3x+4y, where x and y are in metre and U is in joule. If initially particle was at rest, then its speed at t=2s will be
A
10m/s
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B
6m/s
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C
8m/s
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D
7m/s
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Solution
The correct option is A10m/s Given U=3x+4y,
So ∂U∂x=3,∂U∂y=4
We have relation between force and potential energy as →F=−∂U∂x^i−∂U∂y^j=−3^i−4^j |→F|=√32+42=5N
From Newtons second law ma=5 a=5 ms−2(m=1kg)
As acceleration is constant we can apply equations of motion V=u+at Att=2s, V=0+5(2)=10 ms−1.