Question

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 $\overrightarrow{F}=(4-x^2)\hat{i}$ N, where x is in meter and the initial position of the block is x=0.The maximum positive displacement x is

• A. $2\sqrt{3}$ m
• B. 2 m
• C. 4 m
• D. $\sqrt{4}$

Answer 1

The correct option is A $2\sqrt{3}$ m

Maximum booger $=[9x-\frac{x^3}{3}]$

$\Rightarrow 4x-\frac{x^3}{3}=0$

$\Rightarrow x^3=4x+3$

$\Rightarrow x^2=12$

$\Rightarrow x=2\sqrt{3}.$

Hence, the answer is $2\sqrt{3}m.$