Question

decide 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.

a 8170. kg satellite moving at 3540. m/s.

a 7800. kg satellite moving at 3200. m/s.

a 8640. kg satellite moving at 3540. m/s.

a 8170. kg satellite moving at 3200. m/s.

Explanation:

Step1: Recall the kinetic energy formula

The formula for kinetic energy is \(K.E=\frac{1}{2}mv^{2}\), where \(m\) is the mass and \(v\) is the velocity.

Step2: Calculate the kinetic energy for each object

• For the \(8170\ kg\) satellite moving at \(3540\ m/s\): \(K.E_1=\frac{1}{2}\times8170\times(3540)^{2}\)
• For the \(7800\ kg\) satellite moving at \(3200\ m/s\): \(K.E_2=\frac{1}{2}\times7800\times(3200)^{2}\)
• For the \(8640\ kg\) satellite moving at \(3540\ m/s\): \(K.E_3=\frac{1}{2}\times8640\times(3540)^{2}\)
• For the \(8170\ kg\) satellite moving at \(3200\ m/s\): \(K.E_4=\frac{1}{2}\times8170\times(3200)^{2}\)

Since \(\frac{1}{2}\) is a common factor, we can compare \(m\times v^{2}\) values:

• \(m_1v_1^{2}=8170\times(3540)^{2}\)
• \(m_2v_2^{2}=7800\times(3200)^{2}\)
• \(m_3v_3^{2}=8640\times(3540)^{2}\)
• \(m_4v_4^{2}=8170\times(3200)^{2}\)

We know that when comparing two terms \(a\times b^{2}\) and \(c\times d^{2}\), if \(b > d\), \(a\times b^{2}>a\times d^{2}\) (for \(a> 0\)), and if \(a > c\) and \(b = d\), \(a\times b^{2}>c\times b^{2}\)

The order of \(m\times v^{2}\) (and thus kinetic energy) from lowest to highest:
\(7800\times(3200)^{2}<8170\times(3200)^{2}<8170\times(3540)^{2}<8640\times(3540)^{2}\)

Answer:

A \(7800\ kg\) satellite moving at \(3200\ m/s\): \(1\)
A \(8170\ kg\) satellite moving at \(3200\ m/s\): \(2\)
A \(8170\ kg\) satellite moving at \(3540\ m/s\): \(3\)
A \(8640\ kg\) satellite moving at \(3540\ m/s\): \(4\)