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Chapter 5: Problem 26

You push your couch a distance of \(4.00 \mathrm{~m}\) across the living room floor with a horizontal force of \(200.0 \mathrm{~N}\). The force of friction is \(150.0 \mathrm{~N}\). What is the work done by you, by the friction force, by gravity, and by the net force?

Short Answer

Expert verified
Answer: The work done by the applied force (you) is 800.0 J, the work done by the friction force is -600.0 J, the work done by the force of gravity is 0 J, and the work done by the net force is 200.0 J.

Step by step solution

01

Calculate the work done by the applied force (you)

Using the work formula and the given data, we have: Work = (200.0 N) × (4.00 m) × cos(0) Since cos(0) = 1, Work = 200.0 N × 4.00 m = 800.0 J The work done by the applied force (you) is 800.0 J.
02

Calculate the work done by the friction force

Again, using the work formula and the given data, we have: Work = (150.0 N) × (4.00 m) × cos(180) Since cos(180) = -1, Work = 150.0 N × 4.00 m × -1 = -600.0 J The work done by the friction force is -600.0 J.
03

Calculate the work done by the force of gravity

Since the force of gravity is acting perpendicular to the direction of displacement, the angle between them is 90 degrees. Therefore, we have: Work = Force × Distance × cos(90) Since cos(90) = 0, Work = Force × Distance × 0 = 0 The work done by the force of gravity is 0 J.
04

Calculate the net force acting on the couch

The net force acting on the couch can be determined by subtracting the force of friction from the applied force: Net force = Force_applied - Force_friction = 200.0 N - 150.0 N = 50.0 N The net force acting on the couch is 50.0 N.
05

Calculate the work done by the net force

Now that we have the net force, we can calculate the work done using the work formula: Work = (50.0 N) × (4.00 m) × cos(0) Since cos(0) = 1, Work = 50.0 N × 4.00 m = 200.0 J The work done by the net force is 200.0 J. In summary, the work done by the applied force (you) is 800.0 J, the work done by the friction force is -600.0 J, the work done by the force of gravity is 0 J, and the work done by the net force is 200.0 J.

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Most popular questions from this chapter

An ideal spring has the spring constant \(k=440 \mathrm{~N} / \mathrm{m}\) Calculate the distance this spring must be stretched from its equilibrium position for 25 J of work to be done.

A force has the dependence \(F_{x}(x)=-k x^{4}\) on the displacement \(x\), where the constant \(k=20.3 \mathrm{~N} / \mathrm{m}^{4}\). How much work does it take to change the displacement from \(0.73 \mathrm{~m}\) to \(1.35 \mathrm{~m} ?\)

How much work is done against gravity in lifting a \(6.00-\mathrm{kg}\) weight through a distance of \(20.0 \mathrm{~cm} ?\)

A car of mass \(m\) accelerates from rest along a level straight track, not at constant acceleration but with constant engine power, \(P\). Assume that air resistance is negligible. a) Find the car's velocity as a function of time. b) A second car starts from rest alongside the first car on the same track, but maintains a constant acceleration. Which car takes the initial lead? Does the other car overtake it? If yes, write a formula for the distance from the starting point at which this happens. c) You are in a drag race, on a straight level track, with an opponent whose car maintains a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) Both cars have identical masses of \(1000 . \mathrm{kg} .\) The cars start together from rest. Air resistance is assumed to be negligible. Calculate the minimum power your engine needs for you to win the race, assuming the power output is constant and the distance to the finish line is \(0.250 \mathrm{mi}\)

5.60 A man throws a rock of mass \(m=0.325 \mathrm{~kg}\) straight up into the air. In this process, his arm does a total amount of work \(W_{\text {net }}=115 \mathrm{~J}\) on the rock. Calculate the maximum distance, \(h\), above the man's throwing hand that the rock will travel.
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