Question

A block is moving up an inclined plane of inclination ${60}^o$ with velocity of $20m/s$ and stops after $2.00s$. If $g=10m/s^2$ then the approximate value of coefficient of friction is:

• A. $3$
• B. $3.3$
• C. $0.27$
• D. $0.33$

Answer 1

Answer: (C)

$0.27$

$u=20m/s,v=0t=2s$

apply first equation of motion

$v=u+at$

$0=20+a\times 2$

$a=-10m/s^2$

net acceleration along inclined plane $=$ acceleration due to component of weight $+$ due to friction

$a=g\sin \theta +\mu g\cos \theta$

$10=10\sin {60}^o+\mu 10\cos {60}^o$

by solving you will get

$\mu =0.27$