Question

Two workers load identical refrigerators into identical trucks by different methods. One worker has the
refrigerator lifted upward onto the back of the truck, which is 1.5 m above the ground. The other worker uses a
ramp to slide the refrigerator onto the back of the truck. The ramp is 5.0 m long, and raises the refrigerator 1.5
m above the ground. The amount of work done by both workers is the same: 1800 J. What are the magnitudes
of the forces each worker must exert to load the refrigerators?

Answer 1

To determine the magnitude of the forces exerted by each worker, we can use the concept of work done, which is calculated using the formula:

\[
W = F \cdot d \cdot \cos(\theta)
\]

Where:
- \( W \) = work done (in joules),
- \( F \) = force exerted (in newtons),
- \( d \) = distance over which the force is applied (in meters), and
- \( \theta \) = angle between the force and direction of motion.

1. **Worker Lifting the Refrigerator**:
- The first worker lifts the refrigerator vertically with a height of 1.5 m.
- The distance \( d \) over which the force acts is the vertical distance = 1.5 m.
- The angle \( \theta \) between the force (upwards) and the direction of motion (also upwards) is \( 0^\circ \) (cosine of 0 degrees is 1).

Thus, the formula simplifies to:

\[
W = F \cdot d
\]
\[
1800 \, J = F \cdot 1.5 \, m
\]
Solving for \( F \):
\[
F = \frac{1800 \, J}{1.5 \, m} = 1200 \, N
\]

2. **Worker Using the Ramp**:
- For the second worker, the ramp has a length of 5.0 m while it raises the refrigerator 1.5 m vertically.
- The angle \( \theta \) of the force relative to the direction of motion needs to be determined. The height \( h \) (1.5 m) and the length of the ramp \( L \) (5.0 m) give us information about the incline. Using trigonometry:

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1.5}{5.0}
\]
\[
\theta = \arcsin\left(\frac{1.5}{5.0}\right)
\]

However, since we are interested primarily in the work done along the ramp, we continue with our calculation:
The distance \( d \) in the work done equation will be the length of the ramp, so \( d = 5.0 \, m\) and the force acts along the ramp at angle \(\theta\) to the horizontal.

The work equation becomes:

\[
W = F \cdot d \cdot \cos(\theta)
\]
Rearranging for \( F \):
\[
F = \frac{W}{d \cdot \cos(\theta)}
\]
Here, we want to find \( \cos(\theta) \) using the adjacent side (base of the triangle which can be calculated using Pythagorean theorem). The horizontal distance (base) can be calculated using:

\[
x = \sqrt{(5.0)^2 - (1.5)^2} = \sqrt{25 - 2.25} = \sqrt{22.75} \approx 4.77 \, m
\]

Now we need to find \( \cos(\theta) \):

\[
\cos(\theta) = \frac{x}{L} = \frac{4.77}{5.0} \approx 0.954
\]

Now plugging this back into the equation for \( F \):

\[
F = \frac{1800 \, J}{5.0 \, m \cdot 0.954}
\]
\[
F \approx \frac{1800}{4.77} \approx 377.5 \, N
\]

Thus, the forces exerted by each worker are:

- Force by the first worker (lifting directly): \( \mathbf{1200 \, N} \)
- Force by the second worker (using the ramp): \( \mathbf{377.5 \, N} \)