To determine the amount of kinetic energy the electrician possessed just as he touched the ground, we can use the principle of conservation of energy. When the electrician fell from a height of 100 meters, the potential energy (PE) he had at that height was converted into kinetic energy (KE) just before he reached the ground.
The potential energy at the height can be calculated using the formula:
\[ PE = mgh \]
where:
- \( m = 50 , \text{kg} \) (mass of the electrician),
- \( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 100 , \text{m} \) (height).
Now, plugging in the values:
\[ PE = 50 , \text{kg} \times 9.81 , \text{m/s}^2 \times 100 , \text{m} \]
\[ PE = 50 \times 981 = 49050 , \text{J} \]
As the electrician falls, this potential energy gets converted into kinetic energy just before he hits the ground. According to the conservation of mechanical energy, assuming no air resistance and other losses:
\[ KE = PE \]
Therefore, the kinetic energy (KE) just as the electrician touched the ground would be:
\[ KE = 49050 , \text{J} \]
Conclusion: The amount of kinetic energy the electrician possessed just as his body touched the ground is 49,050 joules. This is because the potential energy he had at the height of 100 meters converted into kinetic energy during the free fall.
1 answer