A ball is dropped from the roof of a building. After 6.00 s, the ball has 23.8 J of kinetic energy. The mechanical energy of the drop is 256 J.

What is the potential energy of the ball at this point?

Hint: [tex]\(E = PE + KE\)[/tex]



Answer :

Let's solve the problem step-by-step:

1. Understanding the given information:
- The kinetic energy (KE) of the ball after 6.00 seconds is [tex]\(23.8 \, \text{J}\)[/tex].
- The total mechanical energy (E) of the ball during the drop is [tex]\(256 \, \text{J}\)[/tex].

2. Applying the principle of conservation of mechanical energy:
- According to the principle of conservation of mechanical energy, the total mechanical energy of a system is the sum of its potential energy (PE) and kinetic energy (KE).

[tex]\[ E = PE + KE \][/tex]

3. Rearranging the formula to solve for potential energy (PE):
- We can find the potential energy by rearranging the formula:

[tex]\[ PE = E - KE \][/tex]

4. Substituting the given values:
- Substitute [tex]\( E = 256 \, \text{J} \)[/tex] and [tex]\( KE = 23.8 \, \text{J} \)[/tex] into the equation:

[tex]\[ PE = 256 \, \text{J} - 23.8 \, \text{J} \][/tex]

5. Performing the subtraction:
- Compute the difference:

[tex]\[ PE = 256 \, \text{J} - 23.8 \, \text{J} = 232.2 \, \text{J} \][/tex]

6. Final result:
- The potential energy (PE) of the ball at this point is:

[tex]\[ PE = 232.2 \, \text{J} \][/tex]

In conclusion, the potential energy of the ball after 6.00 seconds is [tex]\( 232.2 \, \text{J} \)[/tex].