The adjoining figure shows a force of 40, N a | vedclass

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(C) Given: Force $F = 40\, N$,angle $	heta = 30^o$,mass $m = 5\, kg$,$g = 10\, ms^{-2}$.$[1]$ The horizontal component of the force is $F_x = F \cos(

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$2, 3$

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(C) Given: Force $F = 40\, N$,angle $	heta = 30^o$,mass $m = 5\, kg$,$g = 10\, ms^{-2}$.$[1]$ The horizontal component of the force is $F_x = F \cos(30^o) = 40 	imes \frac{\sqrt{3}}{2} = 20\sqrt{3}\, N$. Thus,statement $[1]$ is incorrect as it states $20\, N$.$[2]$ The weight of the body $W = mg = 5 	imes 10 = 50\, N$ always acts vertically downwards. Thus,statement $[2]$ is correct.$[3]$ The vertical component of the applied force is $F_y = F \sin(30^o) = 40 	imes 0.5 = 20\, N$ (upwards). The normal force $N$ is given by $N + F_y = mg$,so $N = 50 - 20 = 30\, N$. The net vertical force on the body is zero because it is in equilibrium in the vertical direction. Statement $[3]$ is incorrect.

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