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A conveyor belt is moving at a constant speed of `2m//s` . A box is grenty dropped on it. Th ecoefficient of friction between them is `mu=0.5` . The distance that the box will move relative to belt before coming to rest on it taking `g=10ms^(-2)` is:A. `1.2m`B. `0.6m`C. zeroD. `0.4m` |
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Answer» Correct Answer - D Force, `F=mumg` Retardation of the block on the belt `a=(F)/(m)=(mumg)/(m)=mug` From, `v^(2)=u^(2)+2as` `0=(2)^(2)-2(mug)s` `s=(4)/(2xx0.5xx10)=0.4m` . |
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