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In the figure the variation of potential...


In the figure the variation of potential energy of a particle of mass `m=2kg` is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis. Which of the following statements are correct about the particle:

A

If it is released at the origin it will move in negative x-axis

B

If it is released at `x=2+triangle` where `trianglerarr0` then its maximum speed will be `5(m)/(s)` and it will perform oscillatory motion

C

If initially `x=-10m` and `vecu=sqrt(6)hati(m)/(s)` then it will cross `x=10m`

D

`x=-5m` and `x=+5` m are unstable equilibrium positions of the particle.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
If the particle is released at the origin, it will try to go in the direction of force. Here `(du)/(dx)` is positive and hence force is negative, as a result it will move towards`-ve`x-axis.
When the particle is released at `x=2+triangle`, it will reach the point of least possible potential energy `(-15J)` where it will have maximum kinetic energy.
`(1)/(2)mv_(max)^2=25impliesv_(max)=5(m)/(s)`
The particle will now perform oscillatory motion between `-15JleUle15J`, because reaching `U=+15J`, the kinetic energy and hence speed becomes zero.
`-15JleUle15J`,`U=+15J`,
In (C ), `E_i=U_i+k_i=15+6=21J`
At `x=10m` , `U_f=20JimpliesK_f=1Jne0impliesx=10m`
`implies` the particle cross `x=10m`.
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