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.
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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 \( F \): \[ F = \frac{1800 , J}{1.5 , m} = 1200 , N \]
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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): \( \mathbf{377.5 , N} \)
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