Question

problem 7: you throw a ball of mass 0.12 kg up in the air at a speed of 14 m/s. find the initial kinetic energy of the ball upon release. use conservation of energy to find the height of the ball when the ball is going at a speed of 5 m/s. describe in words what happens to the ball as it raises in the air using the concepts of kinetic and potential energy.

Explanation:

Step1: Calculate initial kinetic energy

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\). Given \(m = 0.12\space kg\) and \(v = 14\space m/s\), we substitute into the formula:
\(K=\frac{1}{2}\times0.12\times14^{2}\)
\(=\frac{1}{2}\times0.12\times196\)
\( = 11.76\space J\)

Step2: Use conservation of energy to find height

By conservation of energy \(E_{initial}=E_{final}\). The initial energy \(E_{i}=K_{i}=\frac{1}{2}mv_{i}^{2}\), and the final energy \(E_{f}=K_{f}+U_{f}=\frac{1}{2}mv_{f}^{2}+mgh\)
So \(\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}+mgh\)
Cancel out \(m\) (since \(m
eq0\)): \(\frac{1}{2}v_{i}^{2}=\frac{1}{2}v_{f}^{2}+gh\)
Rearrange for \(h\): \(h=\frac{v_{i}^{2}-v_{f}^{2}}{2g}\)
Given \(v_{i} = 14\space m/s\), \(v_{f}=5\space m/s\) and \(g = 9.8\space m/s^{2}\)
\(h=\frac{14^{2}-5^{2}}{2\times9.8}\)
\(=\frac{196 - 25}{19.6}\)
\(=\frac{171}{19.6}\approx8.72\space m\)

Step3: Describe energy transformation

As the ball rises, its kinetic energy \(K=\frac{1}{2}mv^{2}\) decreases because its speed \(v\) decreases. At the same time, its gravitational potential energy \(U = mgh\) increases because its height \(h\) increases. The total mechanical energy \(E=K + U\) is conserved (assuming no air - resistance). When the ball is moving at \(5\space m/s\), it still has some kinetic energy and has also gained a significant amount of potential energy compared to its initial state.

Answer:

• Initial kinetic energy: \(11.76\space J\)
• Height when \(v = 5\space m/s\): Approximately \(8.72\space m\)
• Energy transformation: Kinetic energy decreases, potential energy increases as the ball rises.