Question

describe how the objects in the following table would be listed, if they were listed in order of increasing kinetic energy. that is, select \1\ next to the object with the lowest kinetic energy, select \2\ next to the object with the second lowest kinetic energy, and so on.

Explanation:

Step1: Calculate kinetic energy for each object

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\).

• For the first object (\(m = 74\space kg\), \(v = 6.0\space m/s\)): \(K_1=\frac{1}{2}\times74\times(6.0)^{2}= 1332\space J\)
• For the second object (\(m = 74\space kg\), \(v = 7.6\space m/s\)): \(K_2=\frac{1}{2}\times74\times(7.6)^{2}=\frac{1}{2}\times74\times57.76 = 2137.12\space J\)
• For the third object (\(m = 65\space kg\), \(v = 6.0\space m/s\)): \(K_3=\frac{1}{2}\times65\times(6.0)^{2}= 1170\space J\)
• For the fourth object (\(m = 74\space kg\), \(v = 6.9\space m/s\)): \(K_4=\frac{1}{2}\times74\times(6.9)^{2}=\frac{1}{2}\times74\times47.61= 1761.57\space J\)

Step2: Compare the kinetic energies

We have \(K_3(1170\space J)<K_1(1332\space J)<K_4(1761.57\space J)<K_2(2137.12\space J)\)

Answer:

• A \(74\space kg\) runner moving at \(6.0\space m/s\): \(2\)
• A \(74\space kg\) runner moving at \(7.6\space m/s\): \(4\)
• A \(65\space kg\) runner moving at \(6.0\space m/s\): \(1\)
• A \(74\space kg\) runner moving at \(6.9\space m/s\): \(3\)