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(B) The weight of the block acts downwards: $w = mg = 2 	imes 10 = 20 \,N$.The applied force $F = 90 \,N$ is at an angle of $37^{\circ}$ to the horiz

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$8$

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(B) The weight of the block acts downwards: $w = mg = 2 	imes 10 = 20 \,N$.The applied force $F = 90 \,N$ is at an angle of $37^{\circ}$ to the horizontal.The vertical component of the force is $F_V = F \sin 37^{\circ} = 90 	imes \frac{3}{5} = 54 \,N$ (upwards).The horizontal component of the force is $F_H = F \cos 37^{\circ} = 90 	imes \frac{4}{5} = 72 \,N$ (towards the wall).This horizontal component acts as the normal force $N$ exerted by the wall on the block: $N = 72 \,N$.The maximum frictional force is $f_k = \mu N = 0.25 	imes 72 = 18 \,N$.Since the block is moving upwards, the friction acts downwards.The net force in the vertical direction is $F_{	ext{net}} = F_V - w - f_k = 54 - 20 - 18 = 16 \,N$.Using Newton's second law, $F_{	ext{net}} = ma$, we get $16 = 2 	imes a$.Therefore, $a = 8 \,ms^{-2}$.

$A$ block of mass $2 \,kg$ is being pushed against a wall by a force $F=90 \,N$ as shown in the figure. If the coefficient of friction is $0.25$, then the magnitude of acceleration of the block is (Take $g=10 \,ms^{-2}$, $\sin 37^{\circ}=\frac{3}{5}$). (in $\,ms^{-2}$)

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