Question

A force F = 5t is applied on a block of mass 10 kg kept on rough horizontal surface $(\mu =0.5)$, in rest what is speed of the block at t = 20 s?

• A. 15 m/s
• B. 18 m/s
• C. 2.5 m/s
• D. cannot be calculated

Answer 1

The correct option is C 2.5 m/s

Friction has more max value is $\mu ms=0.5\times 10\times 10=50N$

So at $t=10sec$ the block will start to move

with acceleration $a=\frac{5t-50}{10}m/s^2$

Speed $dv=u+adt$

u= initial speed = 0

$\Rightarrow dv=adt$ ;

;dv is speed gained in small time interval at after t= 10 second

$\Rightarrow {\int }_{v=0}^{v=v_1}dv={\int }_{t=10}^{t=20}adt$

$\Rightarrow v_1=\frac{1}{10}{\int }_{10}^{20}(0.5t-5)dt=\frac{1}{10}{\{\frac{0.5t^2}{2}-5t\}}_{10}^{20}$

$\Rightarrow v_1=\frac{1}{10}\left[0.25\right(20{)}^2-5\left(20\right)-\left\{0.25\right(10{)}^2-5\left(10\right)\}]$

& $v_1=\frac{1}{10}[100-100-\{25-50\}]$

$\Rightarrow v_1=2.5m/s$