To determine the mechanical energy of the coin right before it hits the ground, we need to calculate the potential energy it initially had at the height from which it was dropped. This potential energy will convert to kinetic energy just before the coin hits the ground due to the conservation of energy principle, assuming no energy is lost to air resistance.
The potential energy ([tex]\(PE\)[/tex]) at the height is given by the formula:
[tex]\[ PE = m \cdot g \cdot h \][/tex]
Where:
- [tex]\(m\)[/tex] is the mass of the coin, which is [tex]\(0.005 \, \text{kg}\)[/tex].
- [tex]\(g\)[/tex] is the acceleration due to gravity, which is [tex]\(9.8 \, \text{m/s}^2\)[/tex].
- [tex]\(h\)[/tex] is the height from which the coin is dropped, which is [tex]\(3 \, \text{m}\)[/tex].
Substitute these values into the formula:
[tex]\[ PE = 0.005 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
Performing the multiplication:
[tex]\[ PE = 0.005 \times 9.8 \times 3 \][/tex]
[tex]\[ PE = 0.147 \, \text{J} \][/tex]
Therefore, the potential energy, which converts entirely into kinetic energy just before the coin hits the ground, is [tex]\(0.147 \, \text{J}\)[/tex].
So the mechanical energy of the coin right before it hits the ground is:
C. [tex]\(0.147 \, \text{J}\)[/tex]
