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Chapter 6: Problem 75

A \(51-\mathrm{kg}\) packing crate is pulled across a rough floor with a rope that is at an angle of \(43^{\circ}\) above the horizontal. If the tension in the rope is \(120 \mathrm{~N}\), how much work is done on the crate to move it \(18 \mathrm{~m}\) ?

Short Answer

Expert verified
The work done on the crate is approximately 1580 J.

Step by step solution

01

Understand the Problem

A crate is being pulled with a rope at an angle, and we want to calculate the work done. The relevant quantities are a tension force of 120 N, an angle of 43掳, and a displacement of 18 m.
02

Recall the Work Formula

The work done by a force is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.
03

Break Down the Forces

Here, \( F = 120 \mathrm{~N} \), \( d = 18 \mathrm{~m} \), and the angle \( \theta = 43^{\circ} \). Insert these values into the work formula.
04

Calculate the Work Done

Calculate the work by plugging in the values: \( W = 120 \cdot 18 \cdot \cos(43^{\circ}) \). Using a calculator, find \( \cos(43^{\circ}) \approx 0.7313537 \). Then compute:\( W \approx 120 \times 18 \times 0.7313537 \approx 1580 \mathrm{~J} \).
05

Interpret the Result

The calculation shows that approximately 1580 joules of work are done on the crate as it is pulled across the floor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force Calculation
Force is a fundamental concept in physics that describes the interaction that causes an object to be pushed or pulled. In our scenario, the force we are interested in is the tension in the rope. It's given as 120 Newtons (N). But why is this important?
  • Force determines how much energy is needed to move an object.
  • In calculations involving work, force is a crucial component.
  • The direction of the force affects how much work is done.
The formula to calculate work is:\[ W = F \cdot d \cdot \cos(\theta) \]where \( W \) is the work done, \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement.
In the exercise, the rope pulls the crate with a force of \( 120\,\text{N} \), creating the movement necessary to do the work.
Angle of Force Application
The angle of force application plays a vital role in how much work is done. When force is applied at an angle, not all of it contributes to moving the object horizontally.
  • The smaller the angle, the more force acts in the direction of motion.
  • A 90掳 angle means no force supports horizontal movement.
  • Physics often requires conversion of angles from degrees to radians for calculations. Here, degrees are sufficient.
The angle given in the problem is \( 43^{\circ} \). This angle is essential because it determines the component of the force in the direction of the crate's displacement.
Use the cosine function to find how much of the force actually helps move the object: \( \cos(43^{\circ}) \). It shows the efficiency of the force's application.
Displacement in Physics
Displacement refers to the overall change in position of an object. It is a vector quantity, meaning it has both magnitude and direction.
  • Only the straight-line distance from start to finish counts in displacement.
  • In our scenario, the crate moves \( 18\,\text{m} \) horizontally.
  • Displacement is crucial for calculating work because it directly influences work via its magnitude.
The displacement of \( 18\,\text{m} \) represents the path over which the force is applied. A longer displacement at the same force and angle would mean more work done on the crate. Understanding displacement helps in recognizing how far an object has moved and how effective the work done is in terms of its movement.
Trigonometric Functions in Physics
Trigonometric functions help break down forces into components. When forces are applied at angles, these functions determine how much of the force acts in the desired direction.
  • Cosine is used to find the component of force in the horizontal direction.
  • Sine could be used for vertical components, but it's not relevant in this example.
  • Using a calculator, \( \cos(43^{\circ}) \) is approximately 0.7313537.
Using \( \cos(\theta) \) lets us understand how effective a force is when not applied directly along the line of movement. In the given example, this calculation shows that only about 73% of the 120 N force is contributing to moving the crate horizontally, much of the work done relies on this value.
Trigonometric functions are a staple in physics, making complex interactions comprehensible by translating angles into useful force components.

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Most popular questions from this chapter

Calculate A farmhand pushes a bale of hay 3.9 m across the floor of a barn. She exerts a force of 88 N at an angle of 25 below the horizontal. How much work has she done?

A player passes a \(0.600-\mathrm{kg}\) basketball down court for a fast break. The ball leaves the player's hands with a speed of \(8.30 \mathrm{~m} / \mathrm{s}\) and slows down to \(7.10 \mathrm{~m} / \mathrm{s}\) at its highest point. Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?

Research and write a report on the power that humans produce in everyday life. Include the power produced by the brain when thinking and the heart when resting. Also select several strenuous activities such as trackand-field events, bicycle racing, and swimming. In each case, explain how the power is determined. Make a table to compare the various power outputs.

Think \& Calculate A pitcher accelerates a 0.14-kg hardball from rest to \(25.5 \mathrm{~m} / \mathrm{s}\) in \(0.075 \mathrm{~s}\). (a) How much work does the pitcher do on the ball? (b) What is the pitcher's power output during the pitch? (c) Suppose the ball reaches \(25.5 \mathrm{~m} / \mathrm{s}\) in less than \(0.075 \mathrm{~s}\). Is the power produced by the pitcher in this case more than, less than, or the same as the power found in part (b)? Explain.

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