Question
A 20.0-kg rock is dropped and hits the ground at a speed of 90.0 m/s.
Calculate the rock’s gravitational potential energy before it was dropped. (Ignore the effects of friction.)
1,800 J
18,000 J
81,000 J
8,100 J
Calculate the rock’s gravitational potential energy before it was dropped. (Ignore the effects of friction.)
1,800 J
18,000 J
81,000 J
8,100 J
Answers
GPT-4o mini
To calculate the gravitational potential energy (PE) of the rock before it was dropped, we can use the formula:
\[
PE = mgh
\]
where:
- \( m \) = mass of the rock (20.0 kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height from which the rock was dropped
We know the final speed of the rock when it hits the ground is 90.0 m/s. We can use the kinematic equation to relate the final speed, initial speed (which is 0 m/s since it is dropped), acceleration (which in this case is gravity), and height:
\[
v^2 = u^2 + 2gh
\]
where:
- \( v \) = final velocity (90.0 m/s)
- \( u \) = initial velocity (0 m/s)
- \( h \) = height
- \( g \) = 9.81 m/s²
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[
v^2 = 2gh
\]
Now we can rearrange the equation to solve for height \( h \):
\[
h = \frac{v^2}{2g} = \frac{(90.0 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2}
\]
Calculating \( h \):
\[
h = \frac{8100 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \approx 413.5 \, \text{m}
\]
Now we can calculate the gravitational potential energy using the height we just found:
\[
PE = mgh = 20.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 413.5 \, \text{m}
\]
Calculating \( PE \):
\[
PE \approx 20.0 \times 9.81 \times 413.5 \approx 81,000 \, \text{J}
\]
Thus, the rock's gravitational potential energy before it was dropped is approximately \( 81,000 \, \text{J} \).
The correct answer is:
**81,000 J**
\[
PE = mgh
\]
where:
- \( m \) = mass of the rock (20.0 kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height from which the rock was dropped
We know the final speed of the rock when it hits the ground is 90.0 m/s. We can use the kinematic equation to relate the final speed, initial speed (which is 0 m/s since it is dropped), acceleration (which in this case is gravity), and height:
\[
v^2 = u^2 + 2gh
\]
where:
- \( v \) = final velocity (90.0 m/s)
- \( u \) = initial velocity (0 m/s)
- \( h \) = height
- \( g \) = 9.81 m/s²
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[
v^2 = 2gh
\]
Now we can rearrange the equation to solve for height \( h \):
\[
h = \frac{v^2}{2g} = \frac{(90.0 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2}
\]
Calculating \( h \):
\[
h = \frac{8100 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \approx 413.5 \, \text{m}
\]
Now we can calculate the gravitational potential energy using the height we just found:
\[
PE = mgh = 20.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 413.5 \, \text{m}
\]
Calculating \( PE \):
\[
PE \approx 20.0 \times 9.81 \times 413.5 \approx 81,000 \, \text{J}
\]
Thus, the rock's gravitational potential energy before it was dropped is approximately \( 81,000 \, \text{J} \).
The correct answer is:
**81,000 J**