At time t=0, a force F=alpha t, where t is ti | vedclass

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(B) Let at $t=t_0$, the body leaves the plane. Then, at $t=t_0$, the normal force $N=0$.The vertical component of the force is $F_y = F \sin 45^{\circ

Vedclass · Accepted Answer

$\frac{50 \sqrt{2}}{\alpha} 	ext{ m/s}$

Vedclass · Answer

(B) Let at $t=t_0$, the body leaves the plane. Then, at $t=t_0$, the normal force $N=0$.The vertical component of the force is $F_y = F \sin 45^{\circ} = \alpha t_0 \sin 45^{\circ}$.For the body to break off the plane, the vertical component of the force must balance the weight of the body:$N + \alpha t_0 \sin 45^{\circ} = mg$Since $N=0$ at the moment of breaking off, we have:$\alpha t_0 \sin 45^{\circ} = mg$Given $m = 1 	ext{ kg}$ and $g = 10 	ext{ m/s}^2$:$\alpha t_0 \left(\frac{1}{\sqrt{2}}ight) = 1 	imes 10$$t_0 = \frac{10 \sqrt{2}}{\alpha} 	ext{ s}$.The horizontal component of the force is $F_x = F \cos 45^{\circ} = \alpha t \cos 45^{\circ}$.The acceleration of the body in the horizontal direction is $a = \frac{F_x}{m} = \frac{\alpha t \cos 45^{\circ}}{1} = \frac{\alpha t}{\sqrt{2}}$.The velocity $V$ at time $t_0$ is given by:$V = \int_0^{t_0} a \, dt = \int_0^{t_0} \frac{\alpha t}{\sqrt{2}} \, dt = \frac{\alpha}{\sqrt{2}} \left[ \frac{t^2}{2} ight]_0^{t_0} = \frac{\alpha t_0^2}{2 \sqrt{2}}$.Substituting $t_0 = \frac{10 \sqrt{2}}{\alpha}$:$V = \frac{\alpha}{2 \sqrt{2}} \left( \frac{10 \sqrt{2}}{\alpha} ight)^2 = \frac{\alpha}{2 \sqrt{2}} 	imes \frac{100 	imes 2}{\alpha^2} = \frac{100}{\sqrt{2} \alpha} = \frac{50 \sqrt{2}}{\alpha} 	ext{ m/s}$.

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