Question

Think \& Calculate To clean a floor, a janitor pushes on a mop handle with a force of \(43 \mathrm{~N}\). (a) If the mop handle is at an angle of \(55^{\circ}\) above the horizontal, how much work is required to push the mop \(0.50 \mathrm{~m}\) ? (b) If the angle the mop handle makes with the horizontal is increased to \(65^{\circ}\), does the work done by the janitor increase, decrease, or stay the same? Explain.

Short answer

(a) 12.32 J; (b) Work decreases when angle increases to 65°.

Step-by-step solution

Understand the Concept of Work

Work is calculated as the product of the force component in the direction of the displacement and the distance over which the force is applied. The formula for work is given by: \[ W = F \cdot d \cdot \cos(\theta) \].Where \( F \) is the force applied, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.

Apply the Formula for Work in Part (a)

Given that the force \( F = 43 \, \text{N} \), the distance \( d = 0.50 \, \text{m} \), and the angle \( \theta = 55^\circ \), we can calculate the work done by substituting these values into the formula:\[ W = 43 \, \text{N} \times 0.50 \, \text{m} \times \cos(55^\circ) \].Calculate \( \cos(55^\circ) \) and then determine the work \( W \). The approximate value of \( \cos(55^\circ) \) is 0.5736.

Calculate the Work Done

Substitute \( \cos(55^\circ) = 0.5736 \) into the equation:\[ W = 43 \, \text{N} \times 0.50 \, \text{m} \times 0.5736 \].This gives:\[ W \approx 12.32 \, \text{J} \].Thus, the work done when the angle is \( 55^\circ \) is approximately 12.32 joules.

Analyze the Effect of Increasing Angle to 65° in Part (b)

When the angle is increased to \( 65^\circ \), calculate whether the work done increases, decreases, or stays the same. With the formula \( W = F \cdot d \cdot \cos(\theta) \), note how \( \cos(65^\circ) \approx 0.4226 \) is less than \( \cos(55^\circ) \). Lower \( \cos(\theta) \) results in less work done.

Conclusion for Part (b)

By increasing the angle to \( 65^\circ \), the horizontal component of the force decreases, thus the work done \( W \) decreases. The change in angle reduces the effective force causing the motion along the direction of displacement.

Key concepts

Force Components

To understand how force components work, let's break down what happens when you push a mop with a handle. The force exerted on the mop is not entirely in the horizontal direction—only a part of it is. This is why we talk about force components. The force can be described in two parts:

• Horizontal Component: Calculates how much of the force is pushing the mop forward along the floor.
• Vertical Component: Describes how much of the force is directed vertically, pushing down or lifting up the mop.

When the mop handle is pushed at an angle, not all 43 N of force contributes to moving the mop across the floor. Only the horizontal component of this force does, which is why we focus on it to calculate work.

Angle of Force

The angle of force is crucial because it determines the effectiveness of the force applied in the motion's direction. The angle between the mop handle and the horizontal ground affects the amount of work done.

• A smaller angle means more of the force is used effectively in the horizontal direction.
• A larger angle results in less effective horizontal force, as more force goes into vertical movement.

Calculating work done involves assessing this angle to determine the corresponding force component. That's why changing the angle from 55° to 65° results in less work done, even though the force applied remains the same.

Work Done Calculation

When calculating work, the direction of force relative to the direction of displacement is crucial. The formula for work takes into account these components, specified as:
\[ W = F \cdot d \cdot \cos(\theta) \]

Here:- \( W \) is the work done.- \( F \) is the force applied.- \( d \) is the distance.- \( \cos(\theta) \) is the cosine of the angle of force.

Simply put, this formula calculates how much of the applied force actually contributes to moving the mop across the 0.50 meters of floor it travels. By substituting in the known values, like the angle and distance, this method provides the work done in joules, as seen in our problem's solution.

Cosine Function in Physics

The cosine function is a fundamental part of physics calculations, especially in problems involving angles and directions. Unlike the force, which is unidirectional, the cosine function allows us to find how much of the force is effective in the displacement's direction.

• Cosine helps separate the effective force (horizontal) from the total force applied.
• With angles: \( \cos(\theta) \) decreases as the angle \( \theta \) increases, leading to lesser work done along the desired direction.

In this exercise, \( \cos(55°) \approx 0.5736 \) explains that roughly 57.36% of the total force is used effectively, while \( \cos(65°) \) reduces it further. Understanding this concept helps explain why work decreases with an increase in the angle of force.