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 $5 m/s$. The block $1^{st}$ stops at $a$ distance of .......... $m$ from starting point
Difficult
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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.
$m a=m g \sin 37^{\circ}+\mu N=m g \sin 37^{\circ}+\mu m g \cos 37^{\circ}$
or $a=10 \sin 37^{\circ}+(0.5) 10 \cos 37^{\circ}=10 \mathrm{m} / \mathrm{s}^{2}$
using formula, $v^{2}=u^{2}-2 a S,$ we get
$0=5^{2}-2(10) S \Rightarrow S=\frac{25}{20}=1.25 \mathrm{m}$
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