Question

the 2.0 kg wood box in (figure 1) slides down a vertical wood wall while you push on it at a 45° angle. the coefficient of kinetic friction of wood on wood is μk = 0.200. for help with math skills, you may want to review: vector components for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of ready for takeoff. part a what magnitude of force should you apply to cause the box to slide down at a constant speed? express your answer with the appropriate units.

Explanation:

Step1: Analyze forces in the x - direction

According to Newton's second law \(F_{net,x}=ma_x\). Since the box is moving at a constant speed (\(a_x = 0\)), the sum of forces in the \(x\) - direction is \(F_{N}-F\sin45^{\circ}=0\), so \(F_{N}=F\sin45^{\circ}\).

Step2: Analyze forces in the y - direction

According to Newton's second law \(F_{net,y}=ma_y\). Since \(a_y = 0\) (constant speed), \(mg - F\cos45^{\circ}-f_k=0\). The kinetic - friction force \(f_k=\mu_kF_{N}\). Substitute \(F_{N}=F\sin45^{\circ}\) into the \(y\) - direction force equation: \(mg - F\cos45^{\circ}-\mu_kF\sin45^{\circ}=0\).

Step3: Solve for \(F\)

Factor out \(F\) from the terms involving \(F\) in the \(y\) - direction equation: \(mg=F(\cos45^{\circ}+\mu_k\sin45^{\circ})\). Then \(F=\frac{mg}{\cos45^{\circ}+\mu_k\sin45^{\circ}}\).
Given \(m = 2.0\space kg\), \(g = 9.8\space m/s^{2}\), \(\mu_k=0.200\), and \(\cos45^{\circ}=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707\).

$$
LATEXBLOCK0
$$

Answer:

\(F = 23.1\space N\)