Question

questions 1-4 is based on the following diagram.
f = 150 n (at 60°)
20 kg
μ = 0.6

1. what is the net force acting on the 20-kg block?

a. 150 n
b. 130 n
c. 100 n
d. 75 n
e. 33 n

Explanation:

Step1: Calculate the horizontal component of the applied force

The horizontal component of the force \(F\) is \(F_{x}=F\cos\theta\). Given \(F = 150\ N\) and \(\theta=60^{\circ}\), so \(F_{x}=150\cos60^{\circ}=150\times\frac{1}{2}=75\ N\)

Step2: Calculate the normal force

The normal force \(N\) is \(N = mg - F\sin\theta\). Given \(m = 20\ kg\), \(g = 10\ m/s^{2}\), \(F = 150\ N\) and \(\theta = 60^{\circ}\), so \(N=20\times10-150\sin60^{\circ}=200 - 150\times\frac{\sqrt{3}}{2}\approx200 - 129.9 = 70.1\ N\)

Step3: Calculate the frictional force

The frictional force \(f=\mu N\). Given \(\mu = 0.6\) and \(N\approx70.1\ N\), so \(f = 0.6\times70.1=42.06\ N\)

Step4: Calculate the net force

The net force \(F_{net}=F_{x}-f\). Substituting \(F_{x}=75\ N\) and \(f = 42.06\ N\), we get \(F_{net}=75 - 42.06=32.94\approx33\ N\)

Answer:

E. 33 N