1 Answer

Dsymonds · Answer 1 · 2023-02-18T23:00:14+0000

We will have the following:

a. The gravitational potential energy will be:

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

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

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

$\begin{gathered} 882J=(1)/(2)(15kg)v^2\Rightarrow(1764J)/(15kg)=v^2 \\ \\ \Rightarrow v=(14√(15))/(5)m/s\Rightarrow v\approx10.84m/s \end{gathered}$

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.

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