Question

A time varying force $F=10\sqrt{2}t$ starts acting on the $3\text{kg}$ block kept on a rough horizontal surface $(\mu =0.2)$ at $t=0$. Find the time when the horizontal motion begins.
(Take $g=10{\text{m/s}}^2$)

• A. $\frac{1}{2}\text{s}$
• B. $\frac{3}{2}\text{s}$
• C. $\frac{5}{2}\text{s}$
• D. $\frac{7}{2}\text{s}$

Answer 1

The correct option is A $\frac{1}{2}\text{s}$

FBD of block-


From above diagram,

$N+F\sin {45}^{\circ }=30$

$\therefore N=30-F\sin {45}^{\circ }$

For horizontal equilibrium,

$f=F\cos {45}^{\circ }$

$\Rightarrow \mu N=F\cos {45}^{\circ }$

$\Rightarrow \mu [30-F\sin {45}^{\circ }]=F\cos {45}^{\circ }$

putting the given value

$\Rightarrow 0.2[30-10\sqrt{2}t\times (\frac{1}{\sqrt{2}})]=10\sqrt{2}t\times \frac{1}{\sqrt{2}}$

$\Rightarrow 6-2t=10t$

$\Rightarrow$ $\begin{array}{c}t=\frac{1}{2}\text{s}\end{array}$

Hence, option $\left(a\right)$ is correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{Concept-static friction increases upto}\\ \text{limiting value as horizontal force increases.}\\ \text{And when horizontal force is further}\\ \text{increased it not able to oppose the relative}\\ \text{motion and the block start horizontal motion.}\end{array}$$