Question

A block of mass m = 3 kg is resting over a rough horizontal surface having coefficient of friction $\mu$ = 1/3. The block is pulled to the rigid by applying a force F, inclined at angle ${37}^0$ with the horizontal as shown in Fig. The force increases with time according to law $F=2t$ newton. Calculate its velocity v at t= 10s. (g = 10 $m{s}^{-2}$)

• A. $\frac{25}{2}m{s}^{-1}$
• B. $\frac{25}{3}m{s}^{-1}$
• C. $\frac{25}{1}m{s}^{-1}$
• D. $\frac{25}{6}m{s}^{-1}$

Answer 1

The correct option is B $\frac{25}{3}m{s}^{-1}$

$mg-F\sin {37}^{\circ }=N-\left(i\right)$

$F\cos \theta =\mu N\text{-(ii) where,}\theta ={37}^{\circ }$

Minimum force required to move block

$F=\frac{\mu mg-F\sin {37}^{\circ }}{\cos {37}^{\circ }}$

$\Rightarrow F\cos {37}^{\circ }+F\sin {37}^{\circ }=\mu mg$

$\Rightarrow F=\frac{\mu mg}{\cos {37}^{\circ }+\sin {37}^{\circ }}$

$\therefore F=\frac{10\times 5}{7}=\frac{50}{7}$

Given, $F=2t$

$F$ net $=2t-f_r[f_r=frictionalforce]$

$\Rightarrow m\times a=2t-\frac{50}{7}[t>\frac{25}{7}s]$

$\Rightarrow a=\frac{2t}{3}-\frac{50}{21}$

$\Rightarrow v=\int \frac{2t}{3}dt-\frac{50}{21}dt$

$\Rightarrow {v=\frac{2t^2}{6}-\frac{50}{21}t]}_{25/7}^{10}$

$\therefore v=\frac{25}{3}{ms}^{-1}$