The potential energy (in joules) of a particle of mass 1kg moving in a plane is given by U=3x+4y. The position coordinates of the point being x and y, measured in metres. If the particle is at rest at (6,4); then
A
its acceleration is of magnitude 5m/s2
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B
its speed when it crosses the y−axis is 10m/s
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C
it crosses the y− axis (x=0) at y=−4
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D
it moves in a straight line passing through the origin (0,0)
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Solution
The correct option is C it crosses the y− axis (x=0) at y=−4 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 ⇒∣∣→a∣∣=5m/s2
Let at time ′t′ the particle cross y− axis Sx=12(ax)t2−6=12(−3)t2⇒t=2sec
Along y− direction: Sy=12(−4)(2)2=−8 y−4=−8 ⇒ Particle crosses y-axis at y=−4
At (6,4):U=34J&KE=0
At (0,−4):U=−16J⇒KE−16=34J ⇒12mv2=50⇒v=10m/s