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The potential energy varphi, in joule, o...

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

A

its acceleration is of magnitude `5m//s^(2)`

B

its speed when it crosses the y-axis is `10m//s`

C

it crosses the y-axis `(x=0)` at `y = -4`

D

it moves in a striaght line passing through the origin `(0, 0)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`U = 3x +4y`
`a_(y)=(F_(y))/(m)=(-(del Uf delx))/(m)=-3`
`a_(x) = (F_(y))/(m)=(-(delU//dely))/(m) =-4`
`rArr |a|=5m//s_(2)`
Let at time 't' particle crosses y-axis
Then `-6 = (1)/(2)(-3)t_(2) rArr t = 2sec`
Along y-direction : `Delta y=(1)/(2) (-4)(2)_(2)=-8`
`rArr` particle crosses y-axis at y = -4
At `(6,4)U = 34` & `KE = 0`
At `(0,-4) U = -16 rArr KE = 50`
or `(1)/(2) mv_(2) = 50 rArr v = 10 m//s`
while crossing y-axis `v = 10 m//s`.
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