In the figure shown,all surfaces are friction | vedclass

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(C) For the block not to slip on the wedge,the pseudo force $ma$ must balance the component of gravity along the incline. The condition for no slippin

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$150$

Vedclass · Answer

(C) For the block not to slip on the wedge,the pseudo force $ma$ must balance the component of gravity along the incline. The condition for no slipping is $a = g 	an 	heta$. Given $a = 10\, m/s^2$ and $g = 10\, m/s^2$,we have $	an 	heta = 1$,so $	heta = 45^\circ$.The normal force $N$ acting on the block is $N = mg \cos 	heta + ma \sin 	heta$.Substituting the values: $N = (1)(10) \cos 45^\circ + (1)(10) \sin 45^\circ = 10(1/\sqrt{2}) + 10(1/\sqrt{2}) = 20/\sqrt{2} = 10\sqrt{2}\, N$.The block moves horizontally with the wedge with acceleration $a = 10\, m/s^2$. The displacement $s$ in time $t = \sqrt{3}\, s$ is $s = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 10 	imes (\sqrt{3})^2 = 5 	imes 3 = 15\, m$.The normal force $N$ acts at an angle $	heta = 45^\circ$ with the vertical,so its horizontal component is $N_x = N \sin 	heta = (10\sqrt{2}) \sin 45^\circ = 10\sqrt{2} 	imes (1/\sqrt{2}) = 10\, N$.The work done by the normal force is $W = N_x 	imes s = 10\, N 	imes 15\, m = 150\, J$.

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