question_answer

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

Correct Answer: B

Solution :

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