Question

The 10 kg block is moving to the left with a speed of $1.2m/s$ at time $t=0.$ A force F is applied as shown in the graph. After $0.2$ s the force continues at the $10N$ level. If the coefficient of kinetic friction is ${\mu }_k=\mathrm{0.2.}$ Determine the time t at which the block comes to a stop. (g= $10m/s^2$)

• A. $0.333s$
• B. $0.526s$
• C. $0.165s$
• D. None of the above.

Answer 1

The correct option is A $0.333s$

As the block moves towards left, thus the frictional force acts towards right i.e. in the direction of external force applied.

Friction force $f={\mu }_{k}mg=\left(0.2\right)\left(10\right)\left(10\right)=20N$

For duration $0\le t\le 0.2$ s :

External force $F=20N$

Total force acting on the block $F^{\prime }=F+f=20+20=40N$

Retardation of block $a=-\frac{F^{\prime }}{m}=\frac{-40}{10}=-4m/s^2$

Velocity of block just after $0.2$ s $v=u+at$ where $t=0.2$ s

$\therefore$ $v=1.2-4\times 0.2=0.4m/s^2$ (towards left)

For duration $t\ge 0.2$ s :

External force $F=10N$

Total force acting on the block $F^{\prime }=F+f=10+20=30N$

Retardation of block $a=-\frac{F^{\prime }}{m}=\frac{-30}{10}=-3m/s^2$

Let at $t^{\prime }$ ( after 0.2 s) the block comes to rest i.e. $v^{\prime }=0$

Using $v^{\prime }=v+a{t}^{\prime }$

$\therefore$ $0=0.4-3t^{\prime }$ $⟹t^{\prime }=0.133$ s

Thus total time after which the block comes to rest $T=t+t^{\prime }=0.2+0.133=0.333$ s