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Question

The potential energy φ (in joule) of a particle of mass 1 kg moving in the x -y plane obeys the law φ=3x+4y, where (x,y) are the coordinates of the particle in meter. If the particle is at rest at (6,4) at time t=0 , then

A
the particle has constant acceleration.
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B
the work done by the external forces on the particle till the instant it crosses the x -axis is 25 J
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C
the speed of the particle when it crosses the y -axis is
15 ms1
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D
the coordinates of the particle at time t=4 s are (18,28).
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Solution

The correct option is D the coordinates of the particle at time t=4 s are (18,28).
Since potential energy of the particle is equal to φ this means
x - component of the force acting on the particle is equal to
Fx=dϕdx
and y - component of the force on the particle is equal to
Fy=dϕdy
Hence, force acting on the particle is F=(3^i+4^j) N. It means, force acting on the particle is constant. Hence, the particle moves with constant acceleration. So option (a) is correct.
Acceleration of the particle will be a=Fm=(3^i+4^j)ms2
Its magnitude is a=32+42=5 ms2
Since, the particle was initially at rest at (6,4), position of the particle at any time t is given by
x=6+12axt2=(632t2)m
and y=4+12ayt2=(42t2) m
When the particle crosses the x -axis , y=0
t1=2 s
Displacement of the particle during this time is
s1=12at2=5 m
Hence, work done by the force upto this instant will be
Fs1=5×5 J=25 J
Hence, option (b) is also correct.
The particle crosses y -axis when x=0
Hence, 632t22=0 or t2=2s
Speed of the particle at this instant will be
v=at2=5×2=10ms1
So, option (c) is wrong
At t=4s,x=632(4)2=18 m
and y=42(4)2=28 m
Hence, option (d) is also correct.

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