A force of 750 , N is applied to a block of m | vedclass

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(D) $1$. The component of the gravitational force acting down the plane is $mg \sin 	heta = 102 	imes 9.8 	imes \sin 30^{\circ} = 102 	imes 9.8

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(D) $1$. The component of the gravitational force acting down the plane is $mg \sin 	heta = 102 	imes 9.8 	imes \sin 30^{\circ} = 102 	imes 9.8 	imes 0.5 = 499.8 \, N \approx 500 \, N$.$2$. The applied force $P = 750 \, N$ acts up the plane.$3$. The net external force along the plane is $F_{	ext{net}} = P - mg \sin 	heta = 750 - 500 = 250 \, N$.$4$. The normal reaction is $R = mg \cos 	heta = 102 	imes 9.8 	imes \cos 30^{\circ} = 102 	imes 9.8 	imes 0.866 \approx 865.6 \, N$.$5$. The limiting static friction is $f_{l} = \mu_s R = 0.4 	imes 865.6 = 346.24 \, N$.$6$. Since the net external force $(250 \, N)$ is less than the limiting static friction $(346.24 \, N)$,the block remains in equilibrium.$7$. According to the law of static friction,the actual frictional force adjusts itself to balance the net external force. Therefore,the frictional force is $250 \, N$ acting down the plane.

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