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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 arrow of mass \(m=88 \mathrm{~g}(0.088 \mathrm{~kg})\) is fired from a bow. The bowstring exerts an average force of \(F=110 \mathrm{~N}\) on the arrow over a distance \(d=78 \mathrm{~cm}(0.78 \mathrm{~m})\) Calculate the speed of the arrow as it leaves the bow.

Which of the following is a correct unit of energy? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) \(\mathrm{kg}^{2} \mathrm{~m}^{2} / \mathrm{s}^{2}\) b) \(\mathrm{kg} \mathrm{m}^{2} / \mathrm{s}\) d) \(\mathrm{kg}^{2} \mathrm{~m} / \mathrm{s}^{2}\)

Supppose you pull a sled with a rope that makes an angle of \(30.0^{\circ}\) to the horizontal. How much work do you do if you pull with \(25.0 \mathrm{~N}\) of force and the sled moves \(25.0 \mathrm{~m} ?\)

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by \(5.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy, Calculate the original speeds of the two cars.

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is \(1143.5 \mathrm{~kg}\). The coefficient of static friction between the bed of the truck and the bottom sack in the stack is \(0.372,\) and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to \(56.6 \mathrm{mph}\) in \(22.9 \mathrm{~s}\). The stack of sacks is \(1 \mathrm{~m}\) from the end of the truck bed. Does the stack slide on the truck bed? The coefficient of kinetic friction between the bottom sack and the truck bed is \(0.257 .\) What is the work done on the stack by the force of friction between the stack and the bed of the truck?
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