A block of mass 5 ,kg is pulled by a force F | vedclass

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(B) The forces acting on the block are the applied force $F$ at an angle of $30^{\circ}$ with the horizontal, the weight $mg$ acting downwards, the no

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) The forces acting on the block are the applied force $F$ at an angle of $30^{\circ}$ with the horizontal, the weight $mg$ acting downwards, the normal reaction $N$ acting upwards, and the kinetic friction $f_k$ acting to the left.Resolving the force $F$ into horizontal and vertical components: $F_x = F \cos 30^{\circ}$ and $F_y = F \sin 30^{\circ}$.For vertical equilibrium: $N + F \sin 30^{\circ} = mg \Rightarrow N = mg - F \sin 30^{\circ}$.Given $m = 5 \,kg$, $g = 9.8 \,m/s^2$, and $\mu = 0.1$, we have $N = (5 	imes 9.8) - F \sin 30^{\circ} = 49 - 0.5F$.The kinetic friction is $f_k = \mu N = 0.1(49 - 0.5F) = 4.9 - 0.05F$.The net force in the horizontal direction is $F_{	ext{net}} = F \cos 30^{\circ} - f_k = ma$.Substituting the values: $F(\frac{\sqrt{3}}{2}) - (4.9 - 0.05F) = 5 	imes 3$.$0.866F - 4.9 + 0.05F = 15$.$0.916F = 19.9$.$F = \frac{19.9}{0.916} \approx 21.73 \,N$.This value is closest to $22 \,N$.

$A$ block of mass $5 \,kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is $0.1$, then the force needed to accelerate the block to $3 \,m/s^2$ to the right is close to (in $\,N$)

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