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A body of mass m, having momentum p, is moving on a rough horizontal surface. If it is stopped in a distance x, the coefficient of friction between the body and the surface is given byA. `mu=(p^(2))/(2gm^(2)x)`B. `mu=(p^(2))/(2mgx)`C. `mu=(p)/(2mgx)`D. `mu=(p)/(2gm^(2)x)` |
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Answer» Correct Answer - A Force of friction = `mu`mg. Therefore, retardation a = `mu`mg/m= `mu`g. Also 2ax= `v^(2)or2am^(2)x=m^(2)v^(2)`. But p =mv Therefore, But `2am^(2)x=p^(2)` But a = `mu`g. Therefore, 2`mugm^(2)x=p^(2)` or `mu=(p^(2))/(2gm^(2)x)` |
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