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

Explanation:

Step1: Recall Work - Energy Theorem

The work - energy theorem states that the work done on an object \(W\) is equal to the change in its kinetic energy \(\Delta KE\). The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\), where \(m\) is the mass of the object and \(v\) is its velocity. The change in kinetic energy \(\Delta KE=KE_{final}-KE_{initial}\).

Step2: Identify Initial and Final Velocities

Let the initial velocity \(v_{i} = 1.4\ m/s\) and the final velocity \(v_{f}=- 1.4\ m/s\) (the negative sign indicates the opposite direction). The mass of the ball \(m = 10\ kg\).

Step3: Calculate Initial Kinetic Energy

Using the formula \(KE_{i}=\frac{1}{2}mv_{i}^{2}\), substitute \(m = 10\ kg\) and \(v_{i}=1.4\ m/s\):
\(KE_{i}=\frac{1}{2}\times10\times(1.4)^{2}=\frac{1}{2}\times10\times1.96 = 9.8\ J\)

Step4: Calculate Final Kinetic Energy

Using the formula \(KE_{f}=\frac{1}{2}mv_{f}^{2}\), substitute \(m = 10\ kg\) and \(v_{f}=- 1.4\ m/s\) (note that \((-1.4)^{2}=1.96\)):
\(KE_{f}=\frac{1}{2}\times10\times(- 1.4)^{2}=\frac{1}{2}\times10\times1.96=9.8\ J\)

Step5: Calculate Work Done

Using the work - energy theorem \(W=\Delta KE=KE_{f}-KE_{i}\). Substitute \(KE_{f} = 9.8\ J\) and \(KE_{i}=9.8\ J\):
\(W=9.8 - 9.8=0\ J\)

Answer:

\(0\)