Question

A block slips with a constant velocity on a plane inclined at an angle ${\theta }^0$. The same block is pushed up the plane with an initial velocity $v_0$. The distance covered by the block before coming to rest is-

• A. $\frac{{v_0}^2}{2g\sin \theta }$
• B. $\frac{{v_0}^2}{4g\sin \theta }$
• C. $\frac{{v_0}^2{\sin }^2\theta }{2g}$
• D. $\frac{{v_0}^2{\sin }^2\theta }{4g}$

Answer 1

Answer: (B)

$\frac{{v_0}^2}{4g\sin \theta }$

When the block slips with constant velocity we see that the net force on it is balanced i.e. gravitational force equals the frictional force.
or
$mgsin\theta =\mu mgcos\theta$
or
$\mu =tan\theta$
Now when the block is projected with velocity $v_o$ we have the deceleration acting on it due gravitational force and frictional force.
Total deceleration $=gsin\theta +\mu gcos\theta$
Thus using the equation $v^2=u^2+2as$ we get
$0=v_0^2-2(gsin\theta +\mu gcos\theta )x$
Substituting $\mu =tan\theta$ we get
$x=\frac{v_o^2}{4gsin\theta }$