Question

a net external force is applied to a 4.58 - kg object that is initially at rest. the net force component along the displacement of the object varies with the magnitude of the displacement as shown in the drawing. what is the speed of the object at s = 20.0 m?

Explanation:

Step1: Calculate the work done

The work done \(W\) by a force - displacement graph is the area under the graph.
The area consists of a triangle from \(s = 0\) to \(s=10\space m\) and a rectangle from \(s = 10\space m\) to \(s = 20\space m\).
For the triangle: \(A_{1}=\frac{1}{2}\times base\times height\). Here, base \(b = 10\space m\) and height \(h=10\space N\). So \(A_{1}=\frac{1}{2}\times10\times10 = 50\space J\).
For the rectangle: \(A_{2}=length\times width\). Here, length \(l=(20 - 10)\space m=10\space m\) and width \(w = 10\space N\). So \(A_{2}=10\times10=100\space J\).
The total work \(W=A_{1}+A_{2}=50 + 100=150\space J\).

Step2: Use the work - energy theorem

The work - energy theorem states that \(W=\Delta K=K_{f}-K_{i}\). Since the object is initially at rest, \(K_{i}=\frac{1}{2}mv_{i}^{2}=0\) (where \(m = 4.58\space kg\) and \(v_{i}=0\space m/s\)). And \(K_{f}=\frac{1}{2}mv_{f}^{2}\).
We know \(W=\frac{1}{2}mv_{f}^{2}\). Rearranging for \(v_{f}\), we get \(v_{f}=\sqrt{\frac{2W}{m}}\).
Substitute \(W = 150\space J\) and \(m = 4.58\space kg\) into the formula: \(v_{f}=\sqrt{\frac{2\times150}{4.58}}\).
\(v_{f}=\sqrt{\frac{300}{4.58}}\approx\sqrt{65.5}\approx 8.09\space m/s\).

Answer:

\(8.09\space m/s\)