Question

a ball weighing 10 kg traveling at 1.4 m/s hits and rebounds back off a wall and now travels 1.4m/s in the opposite direction. what is the work done on it? enter the answer check it

Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that \(W=\Delta K = K_{f}-K_{i}\), where \(K=\frac{1}{2}mv^{2}\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is the speed of the object.

Step2: Calculate the initial and final kinetic energies

Given \(m = 10\space kg\), \(v_{i}=1.4\space m/s\), and \(v_{f}=1.4\space m/s\).
The initial kinetic energy \(K_{i}=\frac{1}{2}mv_{i}^{2}=\frac{1}{2}\times10\times(1.4)^{2}= 9.8\space J\).
The final kinetic energy \(K_{f}=\frac{1}{2}mv_{f}^{2}=\frac{1}{2}\times10\times(1.4)^{2}=9.8\space J\).

Step3: Calculate the work done

Using the work - energy theorem \(W = K_{f}-K_{i}\).
Substitute the values of \(K_{i}\) and \(K_{f}\): \(W=9.8 - 9.8\).

Answer:

\(0\space J\)