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.
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To determine the total work done in moving the crate, we need to calculate the work done on each segment of the motion separately and then sum them up.
First, let's calculate the work done in dragging the crate along the smooth level floor. The force applied is 30 N at an angle of 25° to the floor, and the distance moved is 3 m. The formula for work is given by:
Work = Force * Distance * cos(angle)
Work1 = 30 N * 3 m * cos(25°) = 77.73 J (to two decimal places)
Next, let's calculate the work done in pulling the crate up the ramp. The force applied is still 30 N, but now the distance moved is 4 m and the angle is 20° to the horizontal. Using the same formula:
Work2 = 30 N * 4 m * cos(20°) = 111.36 J (to two decimal places)
Finally, let's calculate the work done in dragging the crate along the level platform after going up the ramp. The force applied is still 30 N, the distance moved is 5 m, and the angle is 0° (since the platform is level).
Work3 = 30 N * 5 m * cos(0°) = 150 J
Now, we can find the total work done by summing up the individual work values:
Total Work = Work1 + Work2 + Work3
= 77.73 J + 111.36 J + 150 J
= 339.09 J (to two decimal places)
Therefore, the total work done in moving the crate is approximately 339.09 Joules.
To determine the total work done in moving the crate, we need to calculate the work done in each segment of the motion (dragging along the floor, pulling up the ramp, and dragging along the platform), and then sum them up.
First, let's consider the dragging along the floor. The work done is given by the equation:
Work = Force * Distance * cos(theta)
Where:
Force = 30 N (Given)
Distance = 3 m (Given)
theta = 25 degrees (Given)
Plugging in the values, we have:
Work1 = 30 N * 3 m * cos(25 degrees)
Next, let's consider pulling up the ramp. In this case, we need to consider the component of the force parallel to the ramp, which is given by:
Force_parallel = Force * sin(theta)
Where:
Force = 30 N (Given)
theta = 25 degrees (Given)
Plugging in the values, we have:
Force_parallel = 30 N * sin(25 degrees)
Now, we can calculate the work done when pulling up the ramp using this parallel force. The work done is given by the equation:
Work2 = Force_parallel * Distance * cos(ramp_angle)
Where:
Force_parallel = 30 N * sin(25 degrees) (calculated above)
Distance = 4 m (Given)
ramp_angle = 20 degrees (Given)
Plugging in the values, we have:
Work2 = (30 N * sin(25 degrees)) * 4 m * cos(20 degrees)
Lastly, we can consider the dragging along the platform. The work done is again given by:
Work3 = Force * Distance * cos(theta)
Where:
Force = 30 N (Given)
Distance = 5 m (Given)
theta = 25 degrees (Given)
Plugging in the values, we have:
Work3 = 30 N * 5 m * cos(25 degrees)
Now, we can find the total work done by summing up the individual works:
Total Work = Work1 + Work2 + Work3
You can use a calculator or a computer program to evaluate the trigonometric functions and then perform the calculations to find the total work done.