Problem 13 - University Physics with Modern Physics

Question

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by Paul has magnitude 48.0 N and direction 61.0\(^\circ\) south of west. How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0\(^\circ\) east of north?

Step-by-Step Solution

Answer

The work done by Paul's force is approximately -626.6 J.

Step 1: Understanding the Problem and Provided Values

We need to determine the work done by Paul's force during the crate's displacement. Paul exerts a force of 48.0 N, directed 61.0° south of west, while the crate is moved 12.0 m towards 22.0° east of north. Work done is given by the formula: \[ W = F \cdot d \cdot \cos(\theta) \] where \( F \) is the magnitude of the force, \( d \) is the distance moved, and \( \theta \) is the angle between the force and displacement vectors.

Step 2: Break Down Components of Force

Since the force and the displacement are given in different directions, we first need to express Paul's force in terms of its components. Paul's force vector (\( F_{P} \)) can be broken down to:\[ F_{P,x} = 48.0 \cos(61.0^\circ) \]\[ F_{P,y} = 48.0 \sin(61.0^\circ) \]The direction 'south of west' means the x-component is negative, while the y-component is positive (assuming north is positive y and east is positive x when viewed as a typical Cartesian plane orientation).

Step 3: Calculate Force Components

Calculating each component, we find:\[ F_{P,x} = 48.0 \cos(61.0^\circ) \approx -23.3 \text{ N} \]\[ F_{P,y} = 48.0 \sin(61.0^\circ) \approx -42.0 \text{ N} \]These components represent Paul's force in the west and south directions respectively.

Step 4: Break Down Components of Displacement

Similarly, break the displacement into components. The displacement vector \( \vec{d} \) is 12.0 m at 22.0° east of north:\[ d_{x} = 12.0 \sin(22.0^\circ) \]\[ d_{y} = 12.0 \cos(22.0^\circ) \]Here, positive x means east, and positive y means north.

Step 5: Calculate Displacement Components

Calculating these components, we find:\[ d_{x} = 12.0 \sin(22.0^\circ) \approx 4.5 \text{ m} \]\[ d_{y} = 12.0 \cos(22.0^\circ) \approx 11.1 \text{ m} \]

Step 6: Calculate the Angle Between Force and Displacement

The angle between the force vector \( \vec{F_{P}} \) and displacement vector \( \vec{d} \) can be calculated using the dot product:\[ \vec{F_{P}} \cdot \vec{d} = F_{P,x} \cdot d_{x} + F_{P,y} \cdot d_{y} \]Use these results for magnitude calculations:\[ \vec{F_{P}} \cdot \vec{d} = (-23.3 \times 4.5) + (-42.0 \times 11.1) \approx -626.6 \]

Step 7: Calculate Work Done

The work done can be found utilizing the dot product result:\[ W = \vec{F_{P}} \cdot \vec{d} \approx -626.6 \text{ J} \]The negative sign indicates that the work done by Paul is in the opposite direction of the crate's displacement.

Key Concepts

Force Components

In physics, especially when dealing with vector quantities like force, it's important to decompose these vectors into their components. This makes it easier to work with them since we can handle each component separately.
Force components refer to breaking down a vector force into perpendicular components. If you think of directions on a map:

• The x-component might represent east-west directions
• The y-component might represent north-south directions

In this exercise, Paul's force of 48.0 N is directed 61.0° south of west. To analyze this, we break it into x and y components.To find the x-component (west-east direction), we use:\[ F_{P,x} = F \, \cos(\text{{angle}}) \]Therefore, for Paul, \( F_{P,x} = 48.0 \, \cos(61.0°) \).For the y-component (north-south direction), we have:\[ F_{P,y} = F \, \sin(\text{{angle}}) \]So, \( F_{P,y} = 48.0 \, \sin(61.0°) \).Since the direction is south of west:

• The x-component (F_{P,x}) is negative because it's towards the west
• The y-component (F_{P,y}) is negative as it's towards the south

Understanding these components is crucial for calculating the interaction of forces in multiple directions.

Displacement Vector

A displacement vector defines how far and in what direction an object moves from its initial position. Just like forces, displacement vectors can be broken into components to simplify calculations.In this scenario, Paul's crate moves 12.0 m at an angle of 22.0° east of north.
We break this displacement into:

• The x-component (east-west direction): \( d_{x} = 12.0 \, \sin(22.0°) \).
• The y-component (north-south direction): \( d_{y} = 12.0 \, \cos(22.0°) \).

Since the displacement is east of north:

• The x-component (\(d_{x}\)) is positive, moving towards the east.
• The y-component (\(d_{y}\)) is also positive, indicating motion towards the north.

Displacement vectors help us understand precise motion by showing not just how much something moves, but also where it moves.

Angle Between Vectors

The angle between vectors is a vital aspect when calculating work done, as the direction between the force and displacement directly affects the magnitude of work. In mathematical terms, the work done by a force is the dot product of the force and displacement vectors.The formula for work is:\[ W = F \, d \, \cos(\theta) \]where \( \theta \) is the angle between the force and displacement vectors.To find the angle between the vectors, you can use their components in:\[ \vec{F_{P}} \cdot \vec{d} = F_{P,x} \, d_{x} + F_{P,y} \, d_{y} \]The resulting formula helps derive how much of the force contributes to the displacement in its given direction.A positive result implies that the force aids the displacement, while a negative result shows it does the opposite, which is evident with Paul's force acting against the movement direction of the crate.
When we calculated this, we found:\[ \vec{F_{P}} \cdot \vec{d} \approx -626.6 \text{ J} \]This negative sign confirms that the force had an angle greater than 90° from the displacement, showing force against the opportunity.