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In previous question, if `mu=0.3` the acceleration of the block will be:A. zeroB. `(g)/(10)uarr`C. `(g)/(4)darr`D. `(g)/(5)darr` |
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Answer» Correct Answer - D If `mu=0.3` ltbr. Maximum value of resisting force will be `F_("lim")=muN=0.3mg` But have driving force is `mg//2` . Therefoer the friction will be kinetic nature, the block will slid down. Hence acceleration of the block `a=((mg)/(2)-0.3mg)/(m)=0.2g=(g)/(5)darr` . |
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