Question
Force and work unit test
A person pushes a box he applies a force of 400 newtons and he pushes the box for a distance of z4 meters after he stops pushing the box slides forward another meter how much work was done by the person?
A person pushes a box he applies a force of 400 newtons and he pushes the box for a distance of z4 meters after he stops pushing the box slides forward another meter how much work was done by the person?
Answers
GPT-4o mini
To calculate the work done by the person on the box, we can use the formula for work:
\[
\text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta)
\]
Where:
- Work is measured in joules (J).
- Force is measured in newtons (N).
- Distance is measured in meters (m).
- \( \theta \) is the angle between the force and the direction of motion.
In this scenario, since the person is pushing the box in the same direction that it moves, the angle \( \theta \) is 0 degrees. Therefore, \(\cos(0) = 1\).
Given:
- Force (\( F \)) = 400 N
- Distance (\( d \)) = 4 m
Now we can substitute the values into the work formula:
\[
\text{Work} = 400 \, \text{N} \times 4 \, \text{m} \times \cos(0)
\]
\[
\text{Work} = 400 \, \text{N} \times 4 \, \text{m}
\]
\[
\text{Work} = 1600 \, \text{J}
\]
So, the work done by the person is **1600 joules**.
It is also worth noting that the box sliding an additional meter after the person stops pushing does not contribute to the work done by the person but may involve other forces like friction.
\[
\text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta)
\]
Where:
- Work is measured in joules (J).
- Force is measured in newtons (N).
- Distance is measured in meters (m).
- \( \theta \) is the angle between the force and the direction of motion.
In this scenario, since the person is pushing the box in the same direction that it moves, the angle \( \theta \) is 0 degrees. Therefore, \(\cos(0) = 1\).
Given:
- Force (\( F \)) = 400 N
- Distance (\( d \)) = 4 m
Now we can substitute the values into the work formula:
\[
\text{Work} = 400 \, \text{N} \times 4 \, \text{m} \times \cos(0)
\]
\[
\text{Work} = 400 \, \text{N} \times 4 \, \text{m}
\]
\[
\text{Work} = 1600 \, \text{J}
\]
So, the work done by the person is **1600 joules**.
It is also worth noting that the box sliding an additional meter after the person stops pushing does not contribute to the work done by the person but may involve other forces like friction.