A force F = 10t acts on a block of mass 10 , | vedclass

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(B) The forces acting on the block in the vertical direction are the normal force $N$ (upwards),the vertical component of the applied force $F \sin 37

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$t = 50/3 \, 	ext{s}$

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(B) The forces acting on the block in the vertical direction are the normal force $N$ (upwards),the vertical component of the applied force $F \sin 37^{\circ}$ (upwards),and the weight $mg$ (downwards).The equation for vertical equilibrium is:$N + F \sin 37^{\circ} = mg$$N = mg - F \sin 37^{\circ}$The block loses contact with the surface when the normal force $N$ becomes zero.Setting $N = 0$:$0 = mg - F \sin 37^{\circ}$$mg = F \sin 37^{\circ}$Given $m = 10 \, kg$,$g = 10 \, m/s^2$,$F = 10t$,and $\sin 37^{\circ} = 3/5$:$10 	imes 10 = (10t) 	imes (3/5)$$100 = 6t$$t = 100/6 = 50/3 \, 	ext{s}$

$A$ force $F = 10t$ acts on a block of mass $10 \, kg$ as shown in the figure. Find the time $t$ when the block loses contact with the surface.

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