A block of mass 5 text{ kg} is placed on a ro | vedclass

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(C) Given: Mass $m = 5 	ext{ kg}$,coefficient of friction $\mu = 0.1$,angle of inclination $	heta = 30^\circ$,and $g = 10 	ext{ m/s}^2$.The frictio

Vedclass · Accepted Answer

$\frac{5 \sqrt{3}}{2} 	ext{ N}$

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(C) Given: Mass $m = 5 	ext{ kg}$,coefficient of friction $\mu = 0.1$,angle of inclination $	heta = 30^\circ$,and $g = 10 	ext{ m/s}^2$.The frictional force is $f = \mu N = \mu mg \cos 	heta$.$f = 0.1 	imes 5 	imes 10 	imes \cos 30^\circ = 0.5 	imes 5 	imes \frac{\sqrt{3}}{2} = 1.25 \sqrt{3} 	ext{ N}$.To move the block just up the plane,the force $F_1$ must overcome both the component of gravity acting down the plane and the frictional force acting down the plane:$F_1 = mg \sin 	heta + f = 5 	imes 10 	imes \sin 30^\circ + 1.25 \sqrt{3} = 50 	imes 0.5 + 1.25 \sqrt{3} = 25 + 1.25 \sqrt{3} 	ext{ N}$.To prevent the block from sliding down,the force $F_2$ acts up the plane along with friction,balancing the component of gravity acting down the plane:$F_2 + f = mg \sin 	heta \implies F_2 = mg \sin 	heta - f = 25 - 1.25 \sqrt{3} 	ext{ N}$.Now,calculating the difference:$|F_1| - |F_2| = (25 + 1.25 \sqrt{3}) - (25 - 1.25 \sqrt{3}) = 2.5 \sqrt{3} 	ext{ N}$.

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