Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the work done in pushing a stroller 200 feet along a path by applying a constant force of 12 pounds in the direction of motion.

Step-by-Step Solution

Answer

The work done in pushing the stroller is 2400 pound-feet.

Step 1: Identify related variables

In this case, the force \(F\) is 12 pounds, the displacement \(d\) is 200 feet, and the angle between the force and displacement \(\theta\) is 0 degree.

Step 2: Substitute the values into the formula

Now substitute \(F = 12 pounds\), \(d = 200 feet\), and \(\theta = 0 degree\) into the formula: \(W = F \cdot d \cdot \cos(\theta)\). The cosine of 0 degree is 1, so the formula simplifies to \(W = 12 pounds \cdot 200 feet \cdot 1\).

Step 3: Perform the multiplication

Multiplying the values gives \(W = 12 \cdot 200 = 2400\) pound-feet.

Key Concepts

Constant Force

In physics, when we talk about a constant force, we mean that the force remains the same in both magnitude and direction throughout the movement of an object. For example, if you are pushing a stroller with a force of 12 pounds, this force does not change as you push it along the path.
If the force were to vary, the calculations would become more complex because you would have to account for how the changing force affects the movement over time. However, with a constant force, your calculations simplify, making it easier to determine the work done on the object. This constancy is crucial in easily applying physics formulas, such as the work calculation.

• Constant magnitude means the force does not increase or decrease over time.
• Constant direction implies no changes in the angle or trajectory of the applied force.

Understanding constant force helps students get a clearer picture of scenarios in physics where such conditions simplify problem-solving.

Displacement

Displacement in physics refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In the context of working out the physics problem, the stroller is displaced 200 feet along the path.
To understand displacement better, consider:

• It is not the same as distance. Displacement measures the shortest path between two points, whereas distance is the total path traveled.
• Since it is a vector, it's important to consider both how far the object has moved and in what direction.

For our problem, the stroller moves straight ahead without changing direction, making the displacement straightforward. In problems where the path is not straight, calculating displacement might involve more complex vector calculations. This simplification aids in applying the work formula directly because when the path is straight and in the same direction as the force, it becomes easier to use the actual distance as the displacement.

Cosine of Angle

The cosine of the angle between the applied force and the displacement plays a crucial role in calculating work. The formula for work is expressed as:\[W = F \cdot d \cdot \cos(\theta)\]Where:

• \(W\) is the work done,
• \(F\) is the force applied,
• \(d\) is the displacement,
• \(\theta\) is the angle between the force and the direction of displacement.

Cosine helps adjust the work calculation based on how aligned the force and motion are.
If \(\theta = 0\) degrees, the force is perfectly aligned with the displacement direction, yielding \(\cos(0) = 1\). Thus, all the force contributes to the work done. When the angle changes, \(\cos(\theta)\) becomes less than 1, indicating that some force does not contribute to motion. Like in our problem, understanding how angles affect work helps predict if \(\theta = 90\) degrees or \(\theta = 180\) degrees will have different impacts, with \(\cos(90) = 0\) meaning no work is done as the force is perpendicular to displacement.