The potential energy U in joules of a particle of mass 1kg moving in the x−y plane obeys the law U=3x+4y, where (x,y) are the co-ordinates of the particle in metres. If the particle is released from rest at (6,4) at time t=0, then
A
the particle has zero acceleration
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B
co-ordinates of the particle at t=1s is (2,4.5)
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C
co-ordinates of the particle at t=1s is (4.5,2)
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D
co-ordinates of the particle at t=1s is (2,2)
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Solution
The correct option is C co-ordinates of the particle at t=1s is (4.5,2) Given: U=3x+4y We know that, F=−(dUdx^i+dUdy^j+dUdz^k) ⇒F=−(3^i+4^j)N ⇒F=(−3^i−4^j)N ∴ Acceleration of particle=Fm=(−3^i−4^j)m/s2
Since u=0 (initial velocity) ∴ At t=1sec sx=12axt2=12(−3)×1=−1.5m sy=12ayt2=12(−4)×1=−2m
Therefore, final position of particle (snet)x=6−1.5=4.5m (snet)y=4−2=2m ∴ Position after t=1sec is (4.5,2)m