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A block of mass `4kg` is placed on a rough horizontal plane A time dependent force `F = kt^(2)` acts on the block where `k = 2N//s^(2)` Coefficient of friction `mu = 0.8` force of friction between the block and the plane at `t = 2s` isA. `32N`B. `4N`C. `2N`D. `8N` |
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Answer» Correct Answer - d Maximum force of friction `f_(max) = mu mg = 0.8 xx 4 xx 10 =32N` and applied force at `t =2 s is ` `F = kt^(2) = 2 (2)^(2) = 8 N` The body fails to move Hence the force of friction at `t =2s` is equal to applied force `=8N` . |
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