Question

A block is projected upwards on an inclined plane of inclination ${37}^o$ along the line of greatest slope of $\mu =0.5$ with velocity of 5m/s. The block $1^{st}$ stops at a distance of _____ from starting point.

• A. $1.25m$
• B. $2.5m$
• C. $10m$
• D. $12.5m$

Answer 1

The correct option is A $1.25m$

From FBD, both the forces are acting downward along the inclined plane. Hence, there is deceleration of the block while it moves up the inclined plane and velocity decreases and becomes zero at some distance S from the bottom of the inclined plane.
$ma=mg\sin {37}^o+\mu N=mg\sin {37}^o+\mu mg\cos {37}^o$
or $a=10\sin {37}^o+\left(0.5\right)10\cos {37}^o=10m/s^2$
using formula, $v^2=u^2-2aS$, we get
$0=5^2-2\left(10\right)S\Rightarrow S=\frac{25}{20}=1.25m$