Question

1. a 16 kg object is pushed up a ramp with a 150 n force applied parallel to the ground as shown. if the coefficient of friction is 0.40, what is the acceleration of the object?

Explanation:

Step1: Analyze forces perpendicular to the ramp

The normal force \(N\) balances the components of weight \(mg\) and the applied force \(F\) perpendicular to the ramp.
\(N = mg\cos20^{\circ}+F\sin20^{\circ}\)
Substitute \(m = 16\space kg\), \(g = 9.8\space m/s^{2}\), \(F = 150\space N\)
\(N=16\times9.8\times\cos20^{\circ}+150\times\sin20^{\circ}\)
\(N = 16\times9.8\times0.9397+150\times0.3420\)
\(N=147.28+51.3\)
\(N = 198.58\space N\)

Step2: Calculate frictional force

Frictional force \(f=\mu N\), where \(\mu = 0.40\)
\(f=0.40\times198.58 = 79.43\space N\)

Step3: Analyze forces parallel to the ramp

Using Newton's second law \(F_{net}=ma\)
The net force parallel to the ramp \(F_{net}=F\cos20^{\circ}-mg\sin20^{\circ}-f\)
\(F\cos20^{\circ}=150\times0.9397 = 140.96\space N\)
\(mg\sin20^{\circ}=16\times9.8\times0.3420= 53.5\space N\)
\(F_{net}=140.96 - 53.5-79.43\)
\(F_{net}=7.03\space N\)

Step4: Calculate acceleration

Since \(F_{net}=ma\), then \(a=\frac{F_{net}}{m}\)
\(a=\frac{7.03}{16}=0.44\space m/s^{2}\)

Answer:

The acceleration of the object is \(0.44\space m/s^{2}\)