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A conveyor belt is moving at a constant speed of `2m//s` . A box is gently dropped on it. The coefficient 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. zeroB. `0.4 m`C. `1.2 m`D. `0.6 m` |
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Answer» Correct Answer - B `v^(2) = u^(2) + 2 as = 0 + 2 mu gs` `(2)^(2) = 2 xx 5 xx 10 xx s rArr s = 0.4 m` |
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