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Chapter 6: Problem 5

During a tug-of-war, team A pulls on team B by applying a force of \(1100 \mathrm{~N}\) to the rope between them. How much work does team A do if they pull team B toward them a distance of \(2.0 \mathrm{~m} ?\)

Short Answer

Expert verified
Team A does 2200 Joules of work.

Step by step solution

01

Identify the formula for work

The formula for work done when a force is applied over a distance is given by:\[ W = F \cdot d \cdot \cos(\theta) \]where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance moved in the direction of the force, and \( \theta \) is the angle between the force and the direction of movement. In this problem, because the force and movement are in the same direction, \( \theta = 0 \) degrees.
02

Simplify the formula using given conditions

Since the force is applied in the same direction as the movement, \( \theta = 0 \) degrees, so \( \cos(0) = 1 \). Plugging this into the formula, the equation simplifies to:\[ W = F \cdot d \]
03

Insert the given values

Substitute the given values of force \( F = 1100 \) N and distance \( d = 2.0 \) m into the simplified formula:\[ W = 1100 \, \text{N} \times 2.0 \, \text{m} \]
04

Calculate the work done

Perform the multiplication to find the work done:\[ W = 2200 \, \text{J} \]Therefore, the work done by Team A is 2200 Joules (J).

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Key Concepts

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

Understanding Force
In physics, force is a push or pull upon an object resulting from its interaction with another object. It's a vector, meaning it has both magnitude and direction.
One way to think about force is by considering how it affects motion. It can cause an object to start moving, change its direction, or stop altogether.
For the tug-of-war example, Team A exerts a force of 1100 Newtons (N) on the rope. This force is crucial in calculating the work done, as it directly influences the amount of energy applied.
  • Units: Force is measured in Newtons (N). One Newton is the force required to accelerate a 1 kilogram mass by 1 meter per second squared.
  • Direction matters: Since force is a vector, the direction in which it is applied is important.
Displacement and Its Role
Displacement refers to the change in position of an object. It is a vector quantity, which means it also has direction and magnitude.
In the context of work, displacement refers to how far and in what direction the object moves under the influence of a force. For our problem, Team A pulls the rope a distance of 2.0 meters.
The displacement is key in calculating work because it combines with force to determine the total energy transferred.
  • Vector quality: Just like force, displacement must have direction. Even if a large force is applied, if there鈥檚 no displacement in the direction of that force, no work is done.
  • Units: Measured in meters (m), displacement showcases how far movement occurs.
The Angle of Force
The angle between the force and the direction of displacement is a critical factor in determining the work done.
When this angle is 0 degrees, like in the tug-of-war scenario, the force is completely aligned with the direction of displacement. This maximizes the work done because the entire force contributes to the movement.
  • Cosine component: The formula for work includes the term \( \cos(\theta) \), which accounts for the alignment of force and displacement. When \( \theta = 0 \, \cos(\theta) = 1 \,\) meaning full force contributes to work.
  • Impact of angle: If the force were applied at an angle, less energy would go into moving the object. For instance, at 90 degrees, no work would be done as the movement is perpendicular to the force.
Understanding how these concepts interplay makes it clear how work is calculated and why force, displacement, and the angle between them are all vital components.

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Most popular questions from this chapter

Interactive Solution \(\underline{6.33}\) at presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{~m} / \mathrm{s}\). The building is \(31.0 \mathrm{~m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. (a) What type of energy is changing? (b) Is the work being done by the net external force acting on the skier positive, zero, or negative? Why? (c) How is this work related to the change in the energy of the skier? A 70.3 -kg water-skier has an initial speed of \(6.10 \mathrm{~m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{~m} / \mathrm{s}\). Determine the work done by the net external force acting on the skier.

A \(55.0-\mathrm{kg}\) skateboarder starts out with a speed of \(1.80 \mathrm{~m} / \mathrm{s} .\) He does \(+80.0 \mathrm{~J}\) of work on himself by pushing with his feet against the ground. In addition, friction does \(-265 \mathrm{~J}\) of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{~m} / \mathrm{s}\). (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by Lance Armstrong \((m=75.0)\) is \(6.50 \mathrm{~W}\) per kilogram of his body mass. (a) How much work does he do during a \(135-\mathrm{km}\) race in which his average speed is \(12.0 \mathrm{~m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

You are moving into an apartment and take the elevator to the 6th floor. Does the force exerted on you by the elevator do positive or negative work when the elevator (a) goes up and (b) goes down? Explain your answers. Suppose your weight is \(685 \mathrm{~N}\) and that of your belongings is \(915 \mathrm{~N}\). (a) Determine the work done by the elevator in lifting you and your belongings up to the 6 th floor \((15.2 \mathrm{~m})\) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity? Check to see that your answers are consistent with your answers to the Concept Questions.
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