A block of mass m (initially at rest) is slid | vedclass

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(D) The forces acting on the block along the vertical direction are the vertical component of the applied force $F \cos 	heta$,the gravitational forc

Vedclass · Accepted Answer

$\frac{{mg 	imes \pi }}{{2\left( {1 - \mu } ight)}}$

Vedclass · Answer

(D) The forces acting on the block along the vertical direction are the vertical component of the applied force $F \cos 	heta$,the gravitational force $mg$,and the kinetic friction force $f_k = \mu N$. Since the force $F$ acts at an angle $	heta$ with the vertical,the normal force $N = F \sin 	heta$.The net force in the vertical direction is $F_{	ext{net}} = F \cos 	heta - mg - \mu F \sin 	heta = m \frac{dv}{dt}$.Integrating the equation of motion from $t=0$ to $t_0$ (where $	heta = 	heta_0 t_0 = \frac{\pi}{2}$):$\int_{0}^{v_f} m dv = \int_{0}^{t_0} (F \cos(	heta_0 t) - mg - \mu F \sin(	heta_0 t)) dt$.Since the block starts from rest and comes to rest at $t_0$,the change in velocity is zero:$0 = \int_{0}^{\pi/2} (F \cos 	heta - mg - \mu F \sin 	heta) \frac{d	heta}{	heta_0}$.$0 = \frac{F}{	heta_0} [\sin 	heta]_0^{\pi/2} - \frac{mg}{	heta_0} [	heta]_0^{\pi/2} - \frac{\mu F}{	heta_0} [-\cos 	heta]_0^{\pi/2}$.$0 = \frac{F}{	heta_0} (1) - \frac{mg}{	heta_0} (\frac{\pi}{2}) - \frac{\mu F}{	heta_0} (0 - (-1))$.$0 = F - mg \frac{\pi}{2} - \mu F$.$F(1 - \mu) = \frac{mg \pi}{2}$.$F = \frac{mg \pi}{2(1 - \mu)}$.

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