Question

A block of wood resting on an inclined plane of angle $30{}^{\circ }$ , just starts moving down. If the coefficients of friction is 0. 2, its velocity (in ms-1) after 5 s is: $(g=10m{s}^{-2})$

• A. 12. 75
• B. 16. 35
• C. 18. 25
• D. 20

Answer 1

Answer: (B)

Acceleration of block down the plane.
$a=\frac{\mu mg\cos \theta -mg\sin \theta }{m}$ $=\mu g\cos \theta -g\sin \theta$ $=0.2\times 10\times \cos {30}^o-10\times \sin {30}^o$ $=2\times \frac{\sqrt{3}}{2}-10\times \frac{1}{2}$ $=\sqrt{3}-5$ $=1.73-5$ $=-3.278m/s^2$ From equation of motion, $v=u-at$ Velocity after 5 s $v=0+0.327\times 5$ $=16.35m/s$