Question

A block is projected along a rough horizontal surface with a velocity of $5\text{m/s}$. If the block comes to rest after travelling a distance of $4\text{m}$, then the coefficient of kinetic friction is $(\text{Take}g=10{\text{m/s}}^2)$

• A. $0.25$
• B. $0.31$
• C. $0.21$
• D. $0.20$

Answer 1

The correct option is B $0.31$

Let the mass of the block be $m$. Kinetic friction $f$ will be acting in the direction opposite to the velocity of the block.
$\Rightarrow$Friction acts as a decelerating force on the block.
Friction force $f={\mu }_{k}N={\mu }_{k}mg$
Taking direction of initial velocity as $+ve$ direction,
$a=-\frac{f}{m}=-{\mu }_{k}g...\left(i\right)$


Distance travelled by block before coming to rest is $4\text{m}$.
Applying kinematic equation, $v^2=u^2+2as...\left(ii\right)$
where $v=0,u=+5\text{m/s}$
$a=-{\mu }_{k}g{\text{m/s}}^2,s=+4\text{m}$
Putting all values in Eq. $\left(ii\right)$,
$\Rightarrow 0-(5{)}^2=-2\times 4\times ({\mu }_{k}g)$
$\Rightarrow -25=-8{\mu }_{k}g$
$\therefore {\mu }_k=\frac{25}{8\times 10}=0.31$
Hence, option $\left(b\right)$ is correct.

Alternate solution :

Using work energy theorem we can say that :
Work done by all forces $=$ change in kinetic energy
Here, work is done by friction force only.
$\Rightarrow W=\Delta K.E.$
$-fs=-\frac{1}{2}m{u}^2$
$(\because$ work done by friction is negative and final velocity of block is zero$)$
$\Rightarrow {\mu }_{k}mg\times 4=\frac{1}{2}m{u}^2$
$\Rightarrow {\mu }_k=\frac{5^2}{40\times 2}$
$\Rightarrow {\mu }_k=0.31$