A 1,kg block is being pushed against a wall b | vedclass

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(B) The force $F = 75\,N$ is applied at an angle of $37^{\circ}$ to the horizontal. Resolving the force into components: Horizontal component $N = F \

Vedclass · Accepted Answer

$20$

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(B) The force $F = 75\,N$ is applied at an angle of $37^{\circ}$ to the horizontal. Resolving the force into components: Horizontal component $N = F \cos 37^{\circ} = 75 	imes 0.8 = 60\,N$. This is the normal force exerted by the wall on the block. Vertical component $F_y = F \sin 37^{\circ} = 75 	imes 0.6 = 45\,N$ (acting upwards). The weight of the block is $mg = 1 	imes 10 = 10\,N$ (acting downwards). The force of kinetic friction is $f_k = \mu N = 0.25 	imes 60 = 15\,N$. Since the vertical component of the applied force $(45\,N)$ is greater than the weight $(10\,N)$,the block moves upwards. The friction force acts downwards to oppose the motion. Applying Newton's second law in the vertical direction: $F_y - mg - f_k = ma$ $45 - 10 - 15 = 1 	imes a$ $20 = a$ Therefore,the acceleration $a = 20\,m/s^2$.

$A$ $1\,kg$ block is being pushed against a wall by a force $F = 75\,N$ as shown in the Figure. The coefficient of friction is $0.25.$ The magnitude of acceleration of the block is ........ $m/s^2$. (Take $g = 10\,m/s^2$)

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