Unit Energy Module Work page 1 of 2 The Work Done by a Constant Force in Two Dimensions In two dimensions work is the scalar product of the force applied to an object and the displacement of the object W F i r F r cos If the angle between the force and the displacement is less than 90 work is positive If is greater than 90 work is negative If the force and the displacement are perpendicular no work is done by the force In problems involving work it is useful to draw force diagrams Also be sure to identify the angle between the force and the displacement In two dimensions work is the scalar product of the force applied to an object and the displacement of the object W F i r F r cos Consider the work done on an object pulled along the ground by a force applied at an angle relative to the horizontal You care about the work done along the ground in the horizontal direction Notice that the horizontal component of force is F cos In two dimensions the sign of work depends on the angle between the force applied to the object and the object s displacement If the angle between the force and the displacement is less than 90 work is positive as illustrated in the upper left diagram If the force and the displacement are perpendicular no work is done by the force This is shown in the upper right illustration If is greater than 90 work is negative as illustrated in the lower diagrams www thinkwell com info thinkwell com Copyright 2001 Thinkwell Corp All Rights Reserved 1934 doc rev 03 30 2001 Unit Energy Module Work page 2 of 2 The Work Done by a Constant Force in Two Dimensions A block is slid up a frictionless hill with constant speed The hill is at an angle of relative to the horizontal and its length is L You would like to know the work done by gravity on the block When solving on problems involving work it is often useful to draw a force diagram In this example there are three forces an external force pushing the block up the hill a normal force perpendicular to the hill and the weight of the block You are interested in the weight mg since you care about the work done by gravity The angle between the force mg and the displacement L is 90 Using a trigonometric identity you can show that cos 90 sin The work done by gravity on the block is mgL sin Notice that L sin is the height of the hill h The work done by gravity is always mgh where h is the height that the object is lifted no matter what path the object follows www thinkwell com info thinkwell com Copyright 2001 Thinkwell Corp All Rights Reserved 1934 doc rev 03 30 2001
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