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Calculating Work

Credit: CK-12 Foundation
License: CC BY-NC 3.0

In order for the roller coaster to run down the incline by gravitational attraction, it first must have work done on it towing it up to the top of the hill.

Work

The word work has both an everyday meaning and a specific scientific meaning.  In the everyday use of the word, work would refer to anything, which required a person to make an effort.  In physics, however, work is defined as the force exerted on an object multiplied by the distance the object moves due to that force.

W = Fd

In the scientific definition of the word, if you push against an automobile with a force of 200 N for 3 minutes but the automobile does not move, then you have done NO work.  Multiplying 200 N times 0 meters yields zero work.  If you are holding an object in your arms, the upward force you are exerting is equal to the object’s weight.  If you hold the object until your arms become very tired, you have still done no work.  When you lift an object, you exert a force equal to the object’s weight and the object moves due to that lifting force.  If an object weighs 200. N and you lift it 1.50 meters, then your work is, W = Fd = (200. \ \text{N})(1.50 \ \text{m}) = 300. \ \text{N m} .

One of the units you will see for work is the Newton·meter but since a Newton is also a kilogram·m/s 2 , then a Newton·meter is also kg·m 2 /s 2 .  This unit has also been named the Joule (pronounced Jool) in honor of James Prescott Joule, a nineteen century English physicist.

Example Problem:  A boy lifts a box of apples that weighs 185 N.  The box is lifted a height of 0.800 m.  How much work did the boy do?

Solution:   W = Fd = (185 \ \text{N})(0.800 \ \text{m}) = 148 \ \text{N m} = 148 \ \text{Joules}

Work is done only if a force is exerted in the direction of motion.  If the force and motion are perpendicular to each other, no work is done because there is no motion in the direction of the force. If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done.

Example Problem:  Suppose a 125 N force is applied to a lawnmower handle at an angle of 25° with the ground and the lawnmower moves along the surface of the ground.  If the lawnmower moves 56 m, how much work was done?

Solution:  The solution requires that we determine the component of the force that was in the direction of the motion of the lawnmower.  The component of the force that was pushing down on the ground does not contribute to the work done.

F_{\text{parallel}} = (\text{Force})(\cos 25^\circ) = (125 \ \text{N})(0.906) = 113 \ \text{N}

W = F_{\text{parallel}} d = (113 \ \text{N})(56 \ \text{m}) = 630 \ \text{J}

Summary

  • In physics, work is defined as the force exerted on an object multiplied by the distance the object moves due to that force.
  • The unit for work is called the joule.
  • If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done.

Practice

The following website contains practice questions with answers on the topic of work.

http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section6.rhtml

Review

1.  How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m?

2.  A workman carries some lumber up a staircase.  The workman moves 9.6 m vertically and 22 m horizontally.  If the lumber weighs 45 N, how much work was done by the workman?

3.  A barge is pulled down a canal by a horse walking beside the canal.  If the angle of the rope is 60.0°, the force exerted is 400. N, and the barge is pulled 100. M, how much work did the horse do?

Image Attributions

  1. [1]^ Credit: CK-12 Foundation; License: CC BY-NC 3.0

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