Question

A block is projected with a velocity $20\text{m/s}$ up the incline plane of inclination angle ${60}^{\circ }$ and stops after $2\text{s}$. If $g=10{\text{m/s}}^2$, then the approximate value of coefficient of friction is

• A. $1.3$
• B. $0.83$
• C. $0.27$
• D. $0.38$

Answer 1

The correct option is C $0.27$



From FBD, we can see that the force acting on body is in opposite direction of its velocity hence the retarding force on the body is,

$F=mg\sin {60}^{\circ }+\mu mg\cos {60}^{\circ }=mg(\sin {60}^{\circ }+\mu \cos {60}^{\circ })$

The retardation produced in body is given by,

$a=\frac{F}{m}=g(\sin {60}^{\circ }+\mu \cos {60}^{\circ })$

The body stops after a time interval of $2\text{s}$.

Using the kinematic equation we get,

$v=u+at$

Here $a$ will be substituted with $-ve$ sign opposite direction of $a$ w.r.t direction of velocity.

$\Rightarrow 0=20-2\left[g\right(\sin {60}^{\circ }+\mu \cos {60}^{\circ }\left)\right]$

$\Rightarrow \sin {60}^{\circ }+\mu \cos {60}^{\circ }=1$

$\frac{\sqrt{3}}{2}+\frac{\mu }{2}=1$

$\mu =2-\sqrt{3}=2-1.732$

$\therefore \mu \simeq 0.27$