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In figure, the pulley shown is smooth. T...

In figure, the pulley shown is smooth. The spring and the string are light. Block B slides down from the top along the fixed rough wedge of inclination `theta`. Assuming that the block reaches the end of the wedge, find the speed of the block at the end. Take the coefficient of friction between the block and the wedge to be `mu` and that the spring was relaxed when the block was released from the top of the wedge.

Text Solution

Verified by Experts

The correct Answer is:
`sqrt(2/m[mgh-1/2K((h)/(sintheta))^2-mumghcottheta])`
Here block will loose the gravtitational potential energy. This energy will be divided in three ways.
a. Some part will be converted in the KE of block
b. Some part will go into spring
c. Remaining part will be used to do work against friction.
By energy conservation:
`mgh=1/2mv^2+1/2K((h)/(sintheta))^2+mumgcos theta((h)/(sin theta))`
`impliesv=sqrt(2/m[mgh-1/2K((h)/(sintheta))^2-mumghcottheta])`
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In the figure shown the pulley is smooth . The spring and the string are light. The block 'B' slides down from the top along the fixed rought wedge of inclination theta . Assuming that the block reaches the end of the wedge. Find the speed of the block at the end. Take the coefficient of friction between the block and the wedge to be mu and the spring was relaxed when the block was relased from the top of the wedge.

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Knowledge Check

  • Calculate angle of friction between wedge and block system is at rest M coefficient of friction between wedge and block.

    A
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    B
    `(msin^(2)theta)/(M+mcos^(2) theta)`
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