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A block of mass 10 kg pushed by a force F on a horizontal rough plane moves with an acceleration of 5ms^(-2). When force is doubled, its acceleration becomes 18ms^(-2). Find the coefficient of friction between the block and rough horizontal plane. (g=10 ms^(-2)). |
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Answer» Solution :On a rough horizontal plane, ACCELERATION of a block of MASS .m. is given by`a=(P)/(m)-mu_(k) g ""`….(i) INITIALLY, `P =F , a=5 ms^(-2)` `5=(F)/(10) mu_(k)(10) …..(ii)(because m = 10 kg)` When force is doubled i.e., `P=2F , a = 18 ms^(-2)`. `18=(2F)/(10)-mu_(K)(10)`.....(III) MULTIPLYING Eq. (ii) with 2 and subtracting from Eq.(iii) `8=mu_(k)(10)rArr mu_(k)=(8)/(10)=0.8` |
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