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A block of mass M slides along the sides of bowl as shown in the figure. The walls of the bowl are frictionless and the base has coefficient of friction 0.1, and length 0.5 m. The block is released from the point A which is 0.2 m high as shown in figure. Then the block comes to rest. |
Answer» Solution : From law of conservation of energy mgh `=(1)/(2)mv^(2)` `therefore v^(2)=2(0.2)g=0.4g` `therefore (1)/(2)m(0.4)g=0.1(m)(g)(0.5)+mgh_(1)` `therefore h_(1)=0.15 m` Similarly when it reaches to P again it RISES to 0.1 m. And then COMES to Q and rises to 0.5 m and then FINALLY comes to REST at P. |
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