A cold package of hot dogs slides rightward across a frictionless floor through a distance d = 20.5 cm while three forces are applied to it. Two of the forces are horizontal and have the magnitudes F1 = 5.00 N and F2 = 1.00 N; the third force is angled down by = 65.0° and has the magnitude F3 = 4.00 N.
(a) For the 20.5 cm displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package?
No need to help me on this one. I figured it out. Add W1+W2+W3+Wg(0 J)+Wn(0 J).
What is the horizontal force if a student does 292J of work moving a distance of 2.2m?
To find the net work done on the package, you need to calculate the work done by each individual force and then add them together.
The work done by a force is given by the formula:
Work = Force x Distance x cos(θ)
Where:
- Force is the magnitude of the force applied.
- Distance is the displacement of the object.
- θ is the angle between the force and the direction of motion.
For the given problem, you have three forces:
1) Horizontal force F1 = 5.00 N:
The angle between the force and the direction of motion is 0°, so cos(0°) = 1. The work done by this force is:
W1 = F1 x d x cos(0°)
2) Horizontal force F2 = 1.00 N:
Again, the angle between this force and the direction of motion is 0°, so cos(0°) = 1. The work done by this force is:
W2 = F2 x d x cos(0°)
3) Diagonal force F3 = 4.00 N with an angle of 65.0°:
To calculate the work done by this force, you need to find the horizontal component of the force. It can be calculated as:
F3_horizontal = F3 x cos(65.0°)
Then, you can use this horizontal component, along with the displacement, to calculate the work done:
W3 = F3_horizontal x d
Finally, to find the net work done on the package, you add up the three individual works:
Net work = W1 + W2 + W3
The gravitational and normal forces do no work since they act perpendicular to the direction of motion. Therefore, their work done is zero.
So, the net work done on the package by the forces is given by the sum of W1, W2, and W3.
I'm glad to hear that you were able to figure out the answer to your question! It seems like you have correctly determined that for the 20.5 cm displacement, the net work done on the package by the three applied forces is equal to the sum of the work done by each individual force, which can be expressed as W1 + W2 + W3.
Additionally, you correctly identified that the work done by the gravitational force on the package, denoted as Wg, is 0 J since there is no displacement in the vertical direction. Similarly, the work done by the normal force on the package, denoted as Wn, is also 0 J as the normal force acts perpendicular to the displacement of the package.
Therefore, the net work done on the package by the three applied forces is equal to W1 + W2 + W3 + Wg(0 J) + Wn(0 J).
Well done on solving this problem! If you have any further questions, feel free to ask.