At the instant t = 0,a force F = kt (k is a c | vedclass

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Question

(D) The body rests on a smooth horizontal surface. The forces acting on the body are the gravitational force $mg$ acting downwards,the normal reaction

Vedclass · Accepted Answer

$\frac{mg}{k\,\sin \alpha}$

Vedclass · Answer

(D) The body rests on a smooth horizontal surface. The forces acting on the body are the gravitational force $mg$ acting downwards,the normal reaction $N$ acting upwards,and the applied force $F = kt$ acting at an angle $\alpha$ with the horizontal.Resolving the force $F$ into components,we have the vertical component $F_y = F \sin \alpha = kt \sin \alpha$ acting upwards.The equation of motion in the vertical direction is $N + F \sin \alpha = mg$.The body leaves the surface when the normal reaction $N$ becomes zero.Setting $N = 0$,we get $kt \sin \alpha = mg$.Solving for time $t$,we obtain $t = \frac{mg}{k \sin \alpha}$.

At the instant $t = 0$,a force $F = kt$ ($k$ is a constant) acts on a small body of mass $m$ resting on a smooth horizontal surface. The force is applied at an angle $\alpha$ with the horizontal. The time at which the body leaves the surface is:

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