Problem 42 - Precalculus

Question

To slide a crate across the floor, a force of 800 pounds at a \(20^{\circ}\) angle is needed. How much work is done if the crate is dragged 50 feet?

Step-by-Step Solution

Answer

37,588 foot-pounds of work is done.

Step 1: Identify the Key Information

First, we identify the essential information given in the problem. The force applied is 800 pounds at an angle of \(20^{\circ}\), and the distance the crate is dragged is 50 feet.

Step 2: Understand the Work Formula

Work is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.

Step 3: Substitute the Values

Substitute the given values into the work formula. Here, \( F = 800 \) pounds, \( d = 50 \) feet, and \( \theta = 20^{\circ} \). Therefore, the formula becomes: \[ W = 800 \times 50 \times \cos(20^{\circ}) \].

Step 4: Calculate the Cosine of the Angle

Using a calculator, find \( \cos(20^{\circ}) \). The cosine of \( 20^{\circ} \) is approximately 0.9397.

Step 5: Compute the Work Done

Now, calculate the work done by plugging in the cosine value: \[ W = 800 \times 50 \times 0.9397 = 37,588 \text{ foot-pounds} \]. This is the amount of work done to slide the crate.

Key Concepts

Force in Work Calculation

In physics, force is a fundamental concept that represents a push or a pull on an object. Here, the force is applied to move a crate across the floor. In our problem, the force applied is 800 pounds. It's important to understand that this force is directed at an angle (20 degrees) from the direction in which the crate is being dragged.

• Force is a vector quantity, meaning it has both magnitude and direction.
• The magnitude of the force (here 800 pounds) is crucial to calculate the work done on an object.
• The direction of the force impacts how much work is effectively done, especially when it's applied at an angle.

To calculate work effectively, only the component of the force that is parallel to the object's displacement (distance dragged) is used. This requires using trigonometry, specifically the cosine component, to project the force onto the direction of motion.

Understanding Distance in Work

Distance is the measure of how far an object is moved in the direction of the applied force. In this scenario, the distance the crate is dragged is 50 feet. This displacement is crucial in the work done formula because work is dependent on the distance over which the force acts.

• The distance acts as a multiplier in the work formula: the greater the distance, the more work is done, assuming the force is constant.
• Distance is a scalar quantity, meaning it has only magnitude and no direction.

The full distance in the direction of force application is always used to calculate work. In practice, this ensures that the physical movement of the object is accounted for in the energy exerted or work done by the force.

Role of Angle in Calculating Work

The angle between the force applied and the direction of motion plays a critical role in determining how much force actually contributes to the work done. In this problem, the angle is given as 20 degrees. To find how much of the force is actually doing useful work (i.e., pulling the crate forward), we use the cosine of this angle.

• The angle affects the calculation through the cosine function, as seen in the formula: \[ W = F \cdot d \cdot \cos(\theta) \]
• Cosine of the angle helps in finding the effective force component acting along the direction of movement.
• A smaller angle means a larger component of the force is used in doing the work, while a larger angle would reduce this component.

Understanding how to properly account for angles ensures that calculations of work done are accurate, especially in situations where forces are not directly parallel to the direction of movement.