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A 1.5-kg block is initially at rest on a...

A `1.5-kg` block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by `vecF=(4-x^2)veciN`, where x is in meter and the initial position of the block is `x=0`.
The maximum positive displacement x is

A

(a) `2sqrt3`

B

(b) `2m`

C

(c) `4m`

D

(d) `sqrt2m`

Text Solution

Verified by Experts

The correct Answer is:
A

KE of the particle at `x=x` is `K=4x-x^3/3`
At maximum displacement, velocity will be zero
`K=0`, i.e., `x=2sqrt3m`
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