A crate is dragged 3m along a smooth level floor by a 30n force applied at 25deg to the floor. Then it is pulled 4m up a ramp inclined at 20deg to the horizontal, using the same force. Then, the crate is dragged a further 5m along a level platform using the same force. Determine the total work done in moving the crate.
To determine the total work done in moving the crate, we need to calculate the work done in each specific segment of the motion and then add them up.
Let's break down the motion into three segments:
1. Dragging the crate 3m along a smooth level floor.
2. Pulling the crate 4m up a ramp inclined at 20 degrees to the horizontal.
3. Dragging the crate a further 5m along a level platform.
Segment 1 - Dragging along a smooth level floor:
We know that work done (W) is equal to force (F) multiplied by displacement (d) along the direction of the force. In this case, the force applied is 30N.
The component of the force along the direction of displacement is given by F * cos(theta), where theta is the angle between the force and the displacement.
Theta = 25 degrees
Cos(theta) = cos(25) = 0.9063 (approximately)
The displacement is 3m.
So, work done in this segment = F * cos(theta) * d
= 30N * 0.9063 * 3m
= 81.57 Joules (approximately)
Segment 2 - Pulling up the inclined ramp:
In this segment, we need to consider the work done against gravity in pulling the crate up the ramp.
The force applied is still 30N, but now we need to consider its component perpendicular (Fn) and parallel (Fp) to the inclined ramp.
Fn = F * sin(theta) = 30N * sin(20) = 10.23N (approximately)
Fp = F * cos(theta) = 30N * cos(20) = 28.70N (approximately)
The perpendicular force Fn is responsible for moving the crate vertically against gravity.
The displacement in the vertical direction is 4m.
So, work done in this segment = Fn * d
= 10.23N * 4m
= 40.92 Joules (approximately)
Segment 3 - Dragging along a smooth level platform:
Similar to segment 1, the force applied is still 30N, and the displacement is 5m.
The angle between the force and the displacement is 0 degrees since they are in the same direction.
So, work done in this segment = F * cos(theta) * d
= 30N * 1 * 5m
= 150 Joules
Now, we can find the total work done in moving the crate by summing up the work done in each segment.
Total work done = Work in Segment 1 + Work in Segment 2 + Work in Segment 3
= 81.57 Joules + 40.92 Joules + 150 Joules
= 272.49 Joules (approximately)
Therefore, the total work done in moving the crate is approximately 272.49 Joules.