Question

1. a 450 n force is applied to a 64 kg object as shown. if the coefficient of friction is 0.35, what is the acceleration of the object?

Explanation:

Step1: Resolve the applied force

Resolve the \(450\ N\) force into vertical and horizontal components.
The vertical component \(F_{y}=450\sin25^{\circ}\approx450\times0.4226 = 190.17\ N\)
The horizontal component \(F_{x}=450\cos25^{\circ}\approx450\times0.9063=407.835\ N\)

Step2: Calculate the normal force

The normal force \(N\) is given by \(N = mg+F_{y}\), where \(m = 64\ kg\) and \(g = 9.8\ m/s^{2}\)
\(N=64\times9.8 + 190.17=627.2+190.17 = 817.37\ N\)

Step3: Calculate the frictional force

The frictional force \(f=\mu N\), with \(\mu = 0.35\)
\(f=0.35\times817.37\approx286.08\ N\)

Step4: Apply Newton's second law

According to \(F_{net}=ma\), where \(F_{net}=F_{x}-f\)
\(F_{net}=407.835 - 286.08=121.755\ N\)
Since \(F_{net}=ma\), then \(a=\frac{F_{net}}{m}\)
\(a=\frac{121.755}{64}\approx1.90\ m/s^{2}\)

Answer:

The acceleration of the object is approximately \(1.90\ m/s^{2}\)