Question

You are standing on the roof of the school (10 meters) holding a 1.4 kg watermelon. In order to make it
splatter really well on the ground, you throw it upward with a velocity of 8m/s. Calculate the total energy
of the watermelon.
answers:
137.2 J

44.8 J

182 J

112 J

Answer 1

To calculate the total energy of the watermelon at the moment it is thrown, we need to consider both its gravitational potential energy and its kinetic energy.

1. **Gravitational Potential Energy (PE)**:
\[
PE = mgh
\]
where:
- \( m = 1.4 \, \text{kg} \) (mass of the watermelon)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 10 \, \text{m} \) (height above ground)

Plugging in the values:
\[
PE = 1.4 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 10 \, \text{m} = 137.4 \, \text{J}
\]

2. **Kinetic Energy (KE)**:
\[
KE = \frac{1}{2}mv^2
\]
where:
- \( v = 8 \, \text{m/s} \) (initial velocity given)

Plugging in the values:
\[
KE = \frac{1}{2} \times 1.4 \, \text{kg} \times (8 \, \text{m/s})^2 = \frac{1}{2} \times 1.4 \, \text{kg} \times 64 \, \text{m}^2/\text{s}^2 = 44.8 \, \text{J}
\]

3. **Total Energy (TE)**:
Total Energy is the sum of Potential Energy and Kinetic Energy.
\[
TE = PE + KE
\]
\[
TE = 137.4 \, \text{J} + 44.8 \, \text{J} = 182.2 \, \text{J}
\]

Thus, the total energy of the watermelon is approximately **182 J**.

The closest answer from your options is ***182 J***.