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 kinetic energy of the block between `x=0` and `x=2.0m` is
A.2.33 J B.8.67 J C.5.33 J D.6.67 J

Options

A. (a) `2.33J`
B. (b) `8.67J`
C. (c) `5.33J`
D. (d) `6.67J`

Answer

C

Explanation

From work-energy theorem, kinetic energy of the block at
`x=x is K=underset0oversetx int4-x^2dx=4x-x^3/3`
For K to be minimum, `(dK)/(dx)=0`
or `4-x^2=0` or `x=+-2m`
At `x=+2m`, `(d^2K)/(dx^2)` is negative, i.e., kinetic energy K is maximum and
`K_(max)=4xx2-((2)^3)/(3) ~=5.33J`