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A small block of mass `m` is released from a fixed smooth wedge in figure. Initial point is marked as `A`. Bottom of wedge is marked as `B` and at a point `C` the block stops because the straight part of floor is rough. The friction coefficient of the block with the floor is `:`A. `(h)/(x_(0)`B. `(x_(0))/(h)`C. zeroD. 1 |
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Answer» Correct Answer - A Totol work doen by gravity`=` work done against friction ltbr. `mgh=mu mg. x_(0)` `=mu=(h)/(x_(0))` |
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