Respuesta :

We will have the following:

a. The gravitational potential energy will be:

[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]

So, the gravitational potential energy of C is 882 J.

b. The velocity of C right before it hits the ground will be:

[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]

So, the velocity will be approximately 10.84 m/s.

c.

1. We will have that Eg at the initial position will be: B < C

2. Vfinal upon impact with ground: B = C

3. Ek right before hitting he ground: B < C

f.

1. Eg: A > B

2. V final: A > B

3. Ek: A > B

4. V at 2 meters above the ground: A > B

5. Total energy at 2 m above the ground: A > B.