A block of mass 5, kg is kept on a rough hori | vedclass

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(B) The block is moving to the right. The force $F = 20\sqrt{2}\, N$ acts at an angle of $45^{\circ}$ downwards, pushing the block into the floor.Vert

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(B) The block is moving to the right. The force $F = 20\sqrt{2}\, N$ acts at an angle of $45^{\circ}$ downwards, pushing the block into the floor.Vertical forces: The normal reaction $N$ is given by $N = mg + F \sin 45^{\circ} = 5 	imes 10 + 20\sqrt{2} 	imes (1/\sqrt{2}) = 50 + 20 = 70\, N$.Frictional force: $f_k = \mu N = 0.5 	imes 70 = 35\, N$.Horizontal forces: The horizontal component of the applied force is $F_x = F \cos 45^{\circ} = 20\sqrt{2} 	imes (1/\sqrt{2}) = 20\, N$ (acting to the left).The total retarding force is $F_{net} = f_k + F_x = 35 + 20 = 55\, N$.Retardation $a = F_{net} / m = 55 / 5 = 11\, m/s^2$.Using $v = u - at$, where $u = 33\, m/s$, $a = 11\, m/s^2$, and $t = 3\, s$:$v = 33 - (11 	imes 3) = 33 - 33 = 0\, m/s$.

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