Question

consider the four situations:
a. a 3 - kg ball at rest atop a 5m - tall hill
b. a 4 - kg ball at rest atop a 5m - tall hill
c. a 3 - kg ball moving at 2 m/s atop a 5m - tall hill
d. a 4 - kg ball moving at 2 m/s at ground level

1. rank from greatest to least the potential energy of each ball
2. rank from greatest to least the kinetic energy of each ball
3. rank from the greatest to least the total energy of the ball

Explanation:

Step1: Calculate potential energy

The formula for gravitational potential energy is \(U = mgh\). Assuming \(g = 9.8\ m/s^{2}\) and \(h\) (height of the hill) is \(5\ m\) for cases \(A\), \(B\), \(C\) (ground - level \(h = 0\) for \(D\)).

• For \(A\): \(U_{A}=m_{A}gh=(3\ kg)\times9.8\ m/s^{2}\times5\ m = 147\ J\)
• For \(B\): \(U_{B}=m_{B}gh=(4\ kg)\times9.8\ m/s^{2}\times5\ m=196\ J\)
• For \(C\): \(U_{C}=m_{C}gh=(3\ kg)\times9.8\ m/s^{2}\times5\ m = 147\ J\)
• For \(D\): \(U_{D}=m_{D}gh=(4\ kg)\times9.8\ m/s^{2}\times0\ m = 0\ J\)

Step2: Calculate kinetic energy

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

• For \(A\): \(v_{A} = 0\), so \(K_{A}=\frac{1}{2}m_{A}v_{A}^{2}=0\ J\)
• For \(B\): \(v_{B}=0\), so \(K_{B}=\frac{1}{2}m_{B}v_{B}^{2}=0\ J\)
• For \(C\): \(K_{C}=\frac{1}{2}m_{C}v_{C}^{2}=\frac{1}{2}(3\ kg)(2\ m/s)^{2}=6\ J\)
• For \(D\): \(K_{D}=\frac{1}{2}m_{D}v_{D}^{2}=\frac{1}{2}(4\ kg)(2\ m/s)^{2}=8\ J\)

Step3: Calculate total energy

The total energy \(E = K + U\)

• For \(A\): \(E_{A}=K_{A}+U_{A}=0 + 147=147\ J\)
• For \(B\): \(E_{B}=K_{B}+U_{B}=0 + 196=196\ J\)
• For \(C\): \(E_{C}=K_{C}+U_{C}=6+147 = 153\ J\)
• For \(D\): \(E_{D}=K_{D}+U_{D}=8+0 = 8\ J\)

Answer:

1. Potential energy: \(B>A = C>D\)
2. Kinetic energy: \(D>C>A = B\)
3. Total energy: \(B>C>A>D\)