Figure 7-47 shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance d = 21.5 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.
F1 is directed to the right, and f2 is directed to the left.
(a) For the 21.5 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?
J
To find the net work done on the package, you need to calculate the work done by each force separately and then add them up.
The work done by a force is given by the formula:
Work = Force * Displacement * cos(theta)
Where:
- Force is the magnitude of the force
- Displacement is the distance over which the force is applied
- theta is the angle between the force and the displacement
Let's calculate the work done by each force:
1. For horizontal force F1 = 5.00 N:
The displacement is given as d = 21.5 cm = 0.215 m
Since F1 is directed to the right, the angle theta is 0 degrees.
Work1 = F1 * d * cos(0) = 5.00 N * 0.215 m * cos(0) = 1.075 J
2. For horizontal force F2 = 1.00 N:
Since F2 is directed to the left, the angle theta is 180 degrees.
Work2 = F2 * d * cos(180) = 1.00 N * 0.215 m * cos(180) = -0.215 J (negative sign indicates opposite direction)
3. For the angled force F3 = 4.00 N:
Since F3 is angled downwards by 65.0 degrees, the angle theta is 65.0 degrees.
Work3 = F3 * d * cos(65.0) = 4.00 N * 0.215 m * cos(65.0) = 0.874 J
Now, you can find the net work done by adding the individual works:
Net Work = Work1 + Work2 + Work3
Net Work = 1.075 J + (-0.215 J) + 0.874 J = 1.734 J
Therefore, the net work done on the package by the three applied forces is 1.734 J.