Question

A $1.5kg$ block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of the x-axis is applied to the block. The force is given by $\overrightarrow{F}=(4-x^2)\hat{i}$ N, 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 $x=2.0$ m is

• A. $2.33J$
• B. $8.67J$
• C. $5.33J$
• D. $6.67J$

Answer 1

The correct option is C $5.33J$

$\overrightarrow{F}=(4-x^2)\hat{i}$

$\Rightarrow$ Maximum KE $={\int }_0^{2}Fdx$

$={\int }_0^2(4-x^2)dx$

$={[4x-\frac{x^3}{3}]}_0^{\alpha }$

$=8-\frac{8}{3}$

$=5.33J.$

Hence, the answer is $5.33J.$