Question

A block moving with initial velocity $u=20\text{m/s}$ comes to rest after covering distance of $40m$ on a rough plane. Find out coefficient of kinetic friction. (Take $g=10m/s^2$)

• A. $0.5$
• B. $0.25$
• C. $0.75$
• D. Data incomplete

Answer 1

The correct option is A $0.5$

Given,
$u=20\text{m/s}$
$S=40\text{m}$
Using eq, $v^2=u^2+2as$
$0-(20{)}^2=2a\times 40$
$a=-5{\text{m/s}}^2$
Since no external force except friction is acting on the block, therefore
Frictional force = $ma$
= $5m$
Also, friction force= $\mu mg$
therefore, $\mu mg$ = $5m$
$\mu g=5$
$\mu =\frac{5}{g}=0.5$