Work AP style

Energy Energy: the currency of the universe. Everything has to be “paid for” with energy. Energy can’t be created or destroyed, but it can be transformed from one kind to another and it can be transferred from one object to another.

Doing WORK is one way to transfer energy from one object to another. Work = Force x displacement W = F∙d Unit for work is Newton x meter. One Newton-meter is also called a Joule, J.

Work- the transfer of energy

Work = Force · displacement Work is not done unless there is a displacement. If you hold an object a long time, you may get tired, but NO work was done on the object. If you push against a solid wall for hours, there is still NO work done on the wall.

For work to be done, the displacement of the object must be along the same direction as the applied force. They must be parallel. If the force and the displacement are perpendicular to each other, NO work is done by the force.

For example, in lifting a book, the force exerted by your hands is upward and the displacement is upward- work is done. Similarly, in lowering a book, the force exerted by your hands is still upward, and the displacement is downward. The force and the displacement are STILL parallel, so work is still done. But since they are in opposite directions, now it is NEGATIVE work. F F d d

On the other hand, while carrying a book down the hallway, the force from your hands is vertical, and the displacement of the book is horizontal. Therefore, NO work is done by your hands. Since the book is obviously moving, what force IS doing work??? The static friction force between your hands and the book is acting parallel to the displacement and IS doing work! F d

An object is subject to a force given by F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector) What work was done by this force? First find the displacement, s s =  r = r f – r o = (i + 7j) - (-4i + 3j) = 5i + 4j Then, do the dot product W = F · s (6i – 8j) · (5i + 4j) = 30 – 32 = -2 J of work

Example How much work is done to push a 5 kg cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity? W = Force · displacement Which measurement is parallel to the force- the length of the ramp or the height of the ramp? W = 25 N x 7 m W = 175 J 7 m 3 m F = 25 N

Example How much work is done to carry a 5 kg cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity? W = Force · displacement What force is required to carry the cat? Force = weight of the cat Which is parallel to the Force vector- the length of the ramp or the height? d = height NOT length W = mg x h W = 5 kg x 10 m/s 2 x 3 m W = 150 J 7 m 3 m

If all four ramps are the same height, which ramp would require the greatest amount of work in order to carry an object to the top of the ramp? W = F∙  r For lifting an object, the distance, d, is the height. Because they all have the same height, the work for each is the same!

And,….while carrying yourself when climbing stairs or walking up an incline at a constant velocity, only the height is used to calculate the work you do to get yourself to the top! The force required is your weight! Horizontal component of d Vertical component of d Your Force

How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long? ZERO, because your Force is vertical, but the displacement is horizontal.

Example A boy pushes a lawnmower 20 meters across the yard. If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees, how much work did he do?  F Displacement = 20 m F cos  W = (F cos  )d W = (200 N cos 50˚) 20 m W = 2571 J

NOTE: If while pushing an object, it is moving at a constant velocity, the NET force must be zero. So….. Your applied force must be exactly equal to any resistant forces like friction.  F = ma F A – f = ma = 0

A 5.0 kg box is pulled 6 m across a rough horizontal floor (  = 0.4) with a force of 80 N at an angle of 35 degrees above the horizontal. What is the work done by EACH force exerted on it? What is the NET work done? ●Does the gravitational force do any work? NO! It is perpendicular to the displacement. ● Does the Normal force do any work? No! It is perpendicular to the displacement. ● Does the applied Force do any work? Yes, but ONLY its horizontal component! W F = F A cos  x d = 80cos 35˚ x 6 m = J ● Does friction do any work? Yes, but first, what is the normal force? It’s NOT mg! Normal = mg – F A sin  W f = -f x d = -  N∙ d = -  (mg – F A sin  ∙ d = J ● What is the NET work done? J – 7.47 J = J mg Normal FAFA  f

For work to be done, the displacement of the object must be in the same direction, as the applied force. They must be parallel. If the force and the displacement are perpendicular to each other, NO work is done by the force. So, using vector multiplication, W = F d (this is a DOT product!!) (In many university texts, as well as the AP test, the displacement is given by d =  r, where r is the position vector: displacement = change in position W = F  r If the motion is in only one direction, W = F  x Or W = F  y

An object is subject to a force given by F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector) What work was done by this force? First find the displacement,  r r f – r o = (i + 7j) - (-4i + 3j) =  r = 5i + 4j Then, complete the dot product W = F ·  r (6i – 8j) · (5i + 4j) = 30 – 32 = -2J of work

Varying Forces The rule is… “If the Force varies, you must integrate!” If the force varies with displacement- in other words, Force is a function of displacement, you must integrate to find the work done.

Examples of Integration An object of mass m is subject to a force given by F(x) = 3x 3 N. What is the work done by the force?

Examples of Definite Integration To stretch a NON linear spring a distance x requires a force given by F(x) = 4x 2 – x. What work is done to stretch the spring 2 meters beyond its equilibrium position?

Graphing Force vs. postion If you graph the applied force vs. the position, you can find how much work was done by the force. Work = F·d = “area under the curve”. Total Work = 2 N x 2 m + 3N x 4m = 16 J Area UNDER the x-axis is NEGATIVE work = - 1N x 2m Force, N Position, m F d Net work = 16 J – 2 J = 14 J

The Integral = the area under the curve! If y = f(x), then the area under the f(x) curve from x = a to x = b is equal to