question_answer

A block is gently placed at the top of an inclined plane 6.4m long. Find the time taken by the block to slide down to the bottom of the plane. The plane makes an angle\[{{30}^{\text{o}}}\] with the horizontal. Coefficient of friction between the block and the plane is 0.2. Take  \[g=10\,m/{{s}^{2}}\].

Answer:

                As the block is sliding down the inclined plane, force of friction F acts up the plane, Fig. 3(b).38. If a is acceleration produced in the block, then net force on the block down the plane is \[f=ma=mg\,\sin \text{ }\!\!\theta\!\!\text{ }\,\text{- F}\,\text{=}\,mg\,\sin \text{ }\!\!\theta\!\!\text{ }-\mu mg\,\cos \text{ }\!\!\theta\!\!\text{ }=mg\left( \sin \text{ }\!\!\theta\!\!\text{ }-\mu \cos \text{ }\!\!\theta\!\!\text{ } \right)\] or \[a=g\left( \sin \theta -\mu \cos \theta  \right)=10\left( \sin {{30}^{\text{o}}}-0.2\cos {{30}^{\text{o}}} \right)=10\left( \frac{1}{2}-0.2\frac{\sqrt{3}}{2} \right)\] \[a=5\left( 1-0.2\times 1.732 \right)=5\times 0.6536=3.268\,m/{{s}^{2}}\] Now, \[s=6.4\,m\]; u = 0 as the block starts from rest. t = ? From \[s=ut=\frac{1}{2}a{{t}^{2}}\] \[6.4=0+\frac{1}{2}\times 3.268{{t}^{2}}\] \[t=\sqrt{\frac{2\times 6.4}{3.268}}=1.98s\]