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A block of mass 20 kg is acted upon by a force `F=30N` at an angle `53^@` with horizontal in downward direction as shown. The coefficient of friction between the block and the forizontal surface is `0.2`. The friction force acting on the block by the ground is `(g=10(m)/(s^2)`)A. `40.0 N`B. `30.0 N`C. `18.0 N`D. `44.8 N` |
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Answer» Correct Answer - C (Easy) Max. frictional force `f_(max) = mu N` `= mu (mg + F sin 53^(@))` `= 0.2(20 xx 10 + 30 xx) = 44.8 N` As applied horizontal force is `F cos 53^(@)` `= 18 N lt f_(max)`, friction froce will also be 18 N. |
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