Question
A 3 kg object is dropped from a height of 15 meters. What will be its kinetic energy just before hitting the ground? (Ignore air resistance and use g=9.8 m/s2).
Answers
GPT-4o mini
To find the kinetic energy of the object just before it hits the ground, we can first determine its potential energy at the height from which it is dropped. The potential energy \( PE \) of the object at a height \( h \) is given by the formula:
\[
PE = mgh
\]
where:
- \( m = 3 \, \text{kg} \) (mass of the object),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 15 \, \text{m} \) (height).
Now, we can plug in the values:
\[
PE = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 15 \, \text{m}
\]
Calculating this:
\[
PE = 3 \times 9.8 \times 15 = 441 \, \text{J}
\]
The potential energy at the height will be completely converted into kinetic energy \( KE \) just before the object hits the ground (ignoring air resistance). Thus, the kinetic energy \( KE \) just before hitting the ground will also be:
\[
KE = 441 \, \text{J}
\]
Therefore, the kinetic energy of the object just before hitting the ground is **441 Joules**.
\[
PE = mgh
\]
where:
- \( m = 3 \, \text{kg} \) (mass of the object),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 15 \, \text{m} \) (height).
Now, we can plug in the values:
\[
PE = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 15 \, \text{m}
\]
Calculating this:
\[
PE = 3 \times 9.8 \times 15 = 441 \, \text{J}
\]
The potential energy at the height will be completely converted into kinetic energy \( KE \) just before the object hits the ground (ignoring air resistance). Thus, the kinetic energy \( KE \) just before hitting the ground will also be:
\[
KE = 441 \, \text{J}
\]
Therefore, the kinetic energy of the object just before hitting the ground is **441 Joules**.