Two workers are loading crates onto a pick up truck. Worker 1 lifts his crate straight up from the ground to the truck. Worker 2 slides his crate up a ramp which has frictionless rollers. Which statement is true about the work being done on the crate?
Well, it seems like both workers are putting in some effort to move those crates. But if we're talking about the work being done, the crate lifted straight up by Worker 1 would require more work. Worker 2 may have it a bit easier by sliding the crate up the ramp, utilizing those fancy frictionless rollers. But hey, at least they're both working hard and not clowning around!
To determine which statement is true about the work being done on the crate by the two workers, we need to understand the concept of work in physics.
Work, in the context of physics, is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:
W = F × d × cos(θ)
Where:
- W is the work done
- F is the magnitude of the force applied
- d is the displacement of the object
- θ is the angle between the force and the displacement vectors
Now, let's consider the two workers and analyze their situations:
1. Worker 1 lifting the crate straight up:
In this case, Worker 1 is applying a force on the crate perpendicular to the direction of displacement. Since the crate is being lifted straight up vertically, the displacement is in the same direction as the force. As a result, the angle θ between the force and displacement vectors is 0°, and the cosine of 0° is 1.
Thus, the work done by Worker 1 can be calculated as:
W1 = F1 × d1 × cos(0°)
= F1 × d1 × 1
= F1 × d1
This means that the work done by Worker 1 is given by the product of the force applied (F1) and the displacement (d1).
2. Worker 2 sliding the crate up a ramp with frictionless rollers:
In this case, Worker 2 slides the crate up a ramp with frictionless rollers. Since the ramp is designed to reduce the effort required to move the crate, the force required to move the crate is greatly reduced. Also, the displacement is along the ramp's surface.
Since the ramp has frictionless rollers, the force is applied parallel to the displacement, and the angle θ between the force and displacement vectors is 0°. Again, the cosine of 0° is 1.
Therefore, the work done by Worker 2 can be calculated as:
W2 = F2 × d2 × cos(0°)
= F2 × d2 × 1
= F2 × d2
Similar to Worker 1, the work done by Worker 2 is given by the product of the force applied (F2) and the displacement (d2).
Now, considering the two situations, it is evident that both workers are doing work on the crate. However, the amount of work done by each worker depends on the magnitudes of the force applied and the displacement of the crate.
To determine which statement is true about the work being done, we would need more specific information or measurements about the forces and displacements involved.