A block of mass 0.1 is held against a wall applying a horizontal force of 5N on block. If the coefficient of friction between the block and the wall i

Correct Answer - D `a_("max") = (mu mg)/(m) = mu g`

Correct Answer - C `a_(e)=(F)/(m)uarr` `a_(s)=-gdarr` `a_(s//e)=a_(s)-a_(e)=(F)/(m)+g=(F+Mg)/(M)` `s=ut+(I)/(2)at^(2)` `1=(1)/(2)((F+Mg)/(M))t^(2)` `t=sqrt(2(M)/(F+Mg))`

`W=underset(o)overset(2L//3)int-(muMg)/(L)ldl=-(2)/(9)muMgL`

Correct Answer - A `O^(2)-V_(0)^(2)=2(-mug)xxs` `s=(V_(0)^(2))/(2mug)`

Correct Answer - A `F_("req")=mg sin theta+mu mg cos theta`

Correct Answer - B Angle of repose `=tan^(-1)(mi)=` angle of limmiting friction. For just motion friction is limiting.

Correct Answer - B `f_(s)("max")=muN` depends on `mu` and N and not on mass.

Correct Answer - C Friction force opposes relative motion.

Correct Answer - A::B Minimum required force to overcome friction `(mumg)/sqrt(1+mu^(2))=Bilsin thetarArrB=(mumg)/(ilsqrt(1+mu^(2) sin theta) )rArrB_(min)=(mumg)/(ilsqrt(1+mu^(2)))`

Correct Answer - f=ma