A block is being pulled by a force F on a lon | vedclass

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(B) Let the mass of the block be $m$. The forces acting on the block are the gravitational force $mg$ downwards,the normal force $N$ upwards,and the a

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$g \cot 	heta$

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(B) Let the mass of the block be $m$. The forces acting on the block are the gravitational force $mg$ downwards,the normal force $N$ upwards,and the applied force $F$ at an angle $	heta$ with the horizontal.For the block to just lift off the floor,the normal force $N$ must become zero. At this instant,the vertical component of the force $F$ balances the weight of the block:$F \sin 	heta = mg$$F = \frac{mg}{\sin 	heta}$Immediately after the block leaves the floor,the only forces acting on it are the gravitational force $mg$ downwards and the applied force $F$ at an angle $	heta$ with the horizontal.The net force in the horizontal direction is $F_x = F \cos 	heta$.Using Newton's second law,the horizontal acceleration $a_x$ is:$a_x = \frac{F \cos 	heta}{m} = \frac{mg}{\sin 	heta} \cdot \frac{\cos 	heta}{m} = g \cot 	heta$The net force in the vertical direction is $F_y = F \sin 	heta - mg$. Since $F \sin 	heta = mg$,the vertical acceleration $a_y = 0$.Thus,the total acceleration of the block immediately after it leaves the floor is $g \cot 	heta$ in the horizontal direction.

$A$ block is being pulled by a force $F$ on a long frictionless level floor. The magnitude of the force is gradually increased from zero until the block lifts off the floor. Immediately after the block leaves the floor,its acceleration is

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