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Question

The potential energy of a particle of mass 5 kg moving in x-y plane is given as U=7x+24y joule, x and y being in metre. Initially at t=0, the particle is at the origin (0, 0) moving with a velocity of (8.6^i+23.2^j) ms1. Then:

A
the velocity of the particle at t=4 s, is 5 ms1
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B
the acceleration of the particle is 5 ms2
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C
the direction of motion of the particle initially (at t=0) is at right angles to the distance of acceleration
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D
the path of the particle is circle
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Solution

The correct options are
A the velocity of the particle at t=4 s, is 5 ms1
B the acceleration of the particle is 5 ms2
F=[δUδY^i+δUδy^j]=(7^i24^j)N
a=Fm=(75^i245^j) m/s2
^a=(75)2+(245)2=5 m/s2
Since, a= constant, we can apply
v=u+at
=(8.6^i+23.2^j)+(75^i245^j)(4) s
=(3^i+4^j) m/s

v=(3)3+(4)2=5 m/s

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