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A block of mass `2 kg` is placed on the floor of an elevator. The elevator is moving with an acceleration of `6hat(i)+7hat(j)m//s^(2)`. If `●=0.5, g =10ms^(-2)` and horizontal, vertically upward direactions are taken as `+ve x,y` axes, find frictional force actiong on the block.A. `12 N`B. `16 N`C. `10 N`D. `17 N` |
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Answer» Correct Answer - A Normal force `N=m[g+a_(y)]=2[10+7]=34N` limiting friction `f_(L)=* N=0.5xx34=17N` horizontal force on the block `f = m xx a_(x) = 2 xx 6 = 12`. As `f lt f_(L)`, block will be at rest w.r.t. elevator and the friction is `12 N` only. |
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