Question
n planet Easy Math where gravity is 10 m/s/s, there is a roller coaster with a 5,000 kg train. How much work is done to lift the train up to the top of the hill 100 m above the station? The work done is ________ J.
Answers
Answered by
GPT-4o mini
To calculate the work done in lifting the roller coaster train to a height of 100 meters, we can use the work-energy principle which states that the work done against gravity is equal to the increase in potential energy.
The formula for gravitational potential energy (PE) is:
\[
PE = mgh
\]
where:
- \(m\) is the mass (in kg),
- \(g\) is the acceleration due to gravity (in m/s²), and
- \(h\) is the height (in m).
Given:
- \(m = 5000 \, \text{kg}\),
- \(g = 10 \, \text{m/s}^2\),
- \(h = 100 \, \text{m}\).
Now, substituting in the values:
\[
PE = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 \times 100 \, \text{m}
\]
\[
PE = 5000 \times 10 \times 100
\]
\[
PE = 5000000 \, \text{J}
\]
Thus, the work done to lift the train up to the top of the hill is **5,000,000 J** (or **5 MJ**).
The formula for gravitational potential energy (PE) is:
\[
PE = mgh
\]
where:
- \(m\) is the mass (in kg),
- \(g\) is the acceleration due to gravity (in m/s²), and
- \(h\) is the height (in m).
Given:
- \(m = 5000 \, \text{kg}\),
- \(g = 10 \, \text{m/s}^2\),
- \(h = 100 \, \text{m}\).
Now, substituting in the values:
\[
PE = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 \times 100 \, \text{m}
\]
\[
PE = 5000 \times 10 \times 100
\]
\[
PE = 5000000 \, \text{J}
\]
Thus, the work done to lift the train up to the top of the hill is **5,000,000 J** (or **5 MJ**).