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Question

The potential energy of a particle of mass 5kg moving in the xy plane is given by U=(7x+24y)J,x and y being in a meter. If the particle starts from an origin, then the speed of the particle at t=2s is:

A
5m/s
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B
14m/s
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C
17.5m/s
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D
10m/s
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Solution

The correct option is C 10m/s
Force along x-direction is =Fx=δuδx=7^i

Force along y- direction is: Fy=δuδ4=24^j

F=7^i+24^j

a=Fm=15(7^i+24^j)

at t=2sec
v=u+at=0+15(7^i+24^j)×2
v=25(7^i+24^j)
|v|=25(49+576
25×25=10m/s.

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