Question

4.19 q: a 15 - kg wagon is pulled to the right across a surface by a tension of 100 newtons at an angle of 30 degrees above the horizontal. a frictional force of 20 newtons to the left acts simultaneously. what is the acceleration of the wagon?

Explanation:

Step1: Resolve the tension force horizontally

The horizontal component of the tension force \(F_{T,x}=F_T\cos\theta\), where \(F_T = 100\space N\) and \(\theta = 30^{\circ}\).
\(F_{T,x}=100\cos30^{\circ}=100\times\frac{\sqrt{3}}{2}\approx86.6\space N\)

Step2: Calculate the net force

The net force \(F_{net}=F_{T,x}-F_f\), where \(F_f = 20\space N\) (frictional force).
\(F_{net}=86.6 - 20=66.6\space N\)

Step3: Use Newton's second law \(F = ma\) to find acceleration

\(a=\frac{F_{net}}{m}\), where \(m = 15\space kg\)
\(a=\frac{66.6}{15}=4.44\space m/s^{2}\)

Answer:

\(4.44\space m/s^{2}\)