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A block `A` of mass `4 kg` is kept on ground. The coefficient of friction between the block and the ground is `0.8`. The external force of magnitude `30 N` is applied parallel to the ground. The resultant force exerted by the ground on the block is `(g = 10 m//s^(2))`A. `40N`B. `30N`C. `0N`D. `50N` |
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Answer» Correct Answer - D `N=mg=40` `(f_(s))_(max)=muN=(0.8)(40)=32` `f_(s)=ext. ` force `=30` `R^(2)=N^(2)+f_(s)^(2)=(50)^(2):.R=50N`. |
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