At the moment t=0, a time-dependent force F=a | vedclass

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(B) When the body leaves the surface, the normal reaction $N=0$.For vertical equilibrium, the upward component of the force must balance the weight of

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$50 \sqrt{2} \,ms^{-1}$

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(B) When the body leaves the surface, the normal reaction $N=0$.For vertical equilibrium, the upward component of the force must balance the weight of the body:$N + F \sin 45^{\circ} = mg$Since $N=0$ at the moment it leaves the plane, we have:$at \sin 45^{\circ} = mg$Substituting the given values ($a=1 \,Ns^{-1}$, $m=1 \,kg$, $g=10 \,ms^{-2}$):$1 \cdot t \cdot \frac{1}{\sqrt{2}} = 1 \cdot 10$$t = 10 \sqrt{2} \,s$Now, the horizontal component of the force provides acceleration $a_x$:$F \cos 45^{\circ} = ma_x$$at \cos 45^{\circ} = m \frac{dv}{dt}$$v = \int_0^t \frac{a}{m} t \cos 45^{\circ} dt = \frac{a \cos 45^{\circ}}{m} \left[ \frac{t^2}{2} \right]_0^{10 \sqrt{2}}$$v = \frac{1 \cdot \frac{1}{\sqrt{2}}}{1} \cdot \frac{(10 \sqrt{2})^2}{2} = \frac{1}{\sqrt{2}} \cdot \frac{200}{2} = \frac{100}{\sqrt{2}} = 50 \sqrt{2} \,ms^{-1}$

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