100 J of kinetic energy but no gravitational potential energy if GPE of the ground is assumed to be zero.
Explanation
Kinetic Energy
The ball carries kinetic energy for being in motion. The size of that energy is given as:
[tex]\text{KE} = \dfrac{1}{2}\; m\cdot v^{2}[/tex]
where
- [tex]m[/tex] is the mass of the object, and
- [tex]v[/tex] is its speed.
SI units:
- [tex]\text{KE}[/tex]: joules.
- [tex]m[/tex]: kilograms.
- [tex]v[/tex]: meters per second.
For this ball:
[tex]\text{KE} = \dfrac{1}{2}\; m\cdot v^{2} = \dfrac{1}{2} \times 2\times 10^{2} = 100 \; \text{J}[/tex].
Gravitational Potential Energy
Whether the ball carries gravitational potential energy depends on the point of zero potential energy. The ball would carry GPE if the point of zero potential energy is chosen underground. However, the question emphasizes that the ball is "on the ground," which implies that the ground is the reference with a GPE of zero.
The gravitational field near the Earth's surface is constant. As a result, GPE is proportional to height relative to the ground (or the point of zero GPE). The ball is on the ground. Its height is zero. As a result, its GPE is also zero.
