A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30° with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is:
Question
A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30° with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is:
Solution 1
To solve this problem, we need to understand the concept of work done. The work done by a force is given by the formula:
Work = Force x Distance x cos(θ)
where:
- Force is the force applied,
- Distance is the distance over which the force is applied, and
- θ is the angle between the force and the direction of motion.
In this case, the force is parallel to the slope, so the angle θ is 0 degrees. The cosine of 0 degrees is 1, so the formula simplifies to:
Work = Force x Distance
The force applied by the man is 80 N and the distance is 5.0 m, so:
Work = 80 N x 5.0 m = 400 J
Therefore, the work done by the man is 400 Joules.
Solution 2
To solve this problem, we need to understand the concept of work done. The work done by a force is given by the formula:
Work = Force x Distance x cos(θ)
where:
- Force is the force applied,
- Distance is the distance over which the force is applied,
- θ is the angle between the force and the direction of motion.
Here, the force applied by the man is parallel to the slope, so the angle θ is 0 degrees. The cosine of 0 degrees is 1, so the formula simplifies to:
Work = Force x Distance
Given in the problem:
- Force = 80 N
- Distance = 5.0 m
Substituting these values into the formula gives:
Work = 80 N x 5.0 m = 400 J
So, the work done by the man is 400 Joules.
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