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 isA. (a) `2sqrt3`B. (b) `2m`C. (c) `4m`D. (d) `sqrt2m`

A `1.5-kg` block is initially at rest on a horizontal frictionless sur

1.	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 isA. (a) `2sqrt3`B. (b) `2m`C. (c) `4m`D. (d) `sqrt2m`
Answer» Correct Answer - 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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