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A aprtical moves on a rough horizontal ground with same initial velocity say `v_(0)`. If `(3//4)th` of its kinetic energy is lost in friction in time `t_(0)` , then coefficient of friction between the partical and the ground is:A. `(v_(0))/(2 gl_(0))`B. `(v_(0))/(4 gl_(0))`C. `(3 v_(0))/(4 gl_(0))`D. `(v_(0))/(gl_(0))` |
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Answer» `3//4^(th)` energy of lost, i.e., `1//4^(th)` kinetic energy is left . Hence, its velocity becomes `v_(0)//2` under a retardation of mmg in time `t_(0)` `(v_(0))/(2) = v_(0) - mu g t_(0)` or `mu = (v_(0))/(2g t_(0))` |
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