Today’s Concepts: Work & Kinetic Energy Physics 211 Lecture 7 Today’s Concepts: Work & Kinetic Energy I need a refresher on how work works (pun intended).

Let’s go back to just F=ma. Please?

Why are we doing this? Simply because it has proven useful.

Why are we doing this? Simply because it has proven useful.

The Dot Product

Work-Kinetic Energy Theorem The work done by force F as it acts on an object that moves between positions r1 and r2 is equal to the change in the object’s kinetic energy:

Work-Kinetic Energy Theorem: 1-D Example If the force is constant and the directions aren’t changing then this is very simple to evaluate: F d car Demo: suitcase, Rope, scale = Fd In this case since cos(0)=1 This is probably what you remember from High School.

ACT A lighter car and a heavier van, each initially at rest, are pushed with the same constant force F. After both vehicles travel a distance d, which of the following statements is true? (Ignore friction) F d car F d van A) They will have the same velocity B) They will have the same kinetic energy C) They will have the same momentum

Concept – very important Derivation – not so important Concept – very important q A force pushing over some distance will change the kinetic energy.

Work done by gravity near the Earth’s surface path mg

Work done by gravity near the Earth’s surface dlN dl1 mg dl1 dy1 dx1 mg dl2

Work done by gravity near the Earth’s surface dlN Dy dl1 mg dl2

Work-Kinetic Energy Theorem If there are several forces acting then W is the work done by the net (total) force: You can just add up the work done by each force

Checkpoint Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. In which case does the object have the biggest net work done on it by all forces during its motion? H Free Fall Frictionless incline String A) Free Fall B) Incline C) String D) All the same

Checkpoint Comments Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. mgh doesn’t matter about path taken Because the string has the largest displacement it has the largest total work done. Be careful, it’s the VERTICAL displacement where work is done by gravity. w = mgy. Since the height is the same for all the objects, the work done is the same in all cases. Since all three objects have same mass and fall down same height, all the work should be same. I’m actually not too sure. I feel like they all would have the same amount of work because they are at the same height; have the same downward acceleration. But they do NOT have the same downward acceleration! Because the object has the most work. I don’t know what this means. Not sure to tell you the truth, the prelecture slide said the normal force doesn’t do anything so I’m assuming they are the same for all examples. WHY is no work done by the normal force? The normal force is perpendicular to the motion and the dot product is zero for perpendicular vectors.

ACT Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. What is the relationship between their speeds when they reach the bottom? H Free Fall Frictionless incline String A) vf > vi > vp B) vf > vp > vi C) vf = vp = vi

Checkpoint / ACT YES! H Demo: rolling balls on tracks Will you do more examples in lecture? YES! H Free Fall Frictionless incline String A) vf > vi > vp B) vf > vp > vi C) vf = vp = vi Demo: rolling balls on tracks

Checkpoint / ACT H Will you do harder examples in lecture? Free Fall Frictionless incline String A) vf > vi > vp B) vf > vp > vi C) vf = vp = vi Wg = mgH D K = 1/2 mv22 Only gravity will do work: Wg = D K

Checkpoint / ACT A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill? v A) WTOT = 0 B) WTOT > 0 C) WTOT < 0 About 25% got this right… 19

Checkpoint A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill? v A) WTOT = 0 B) WTOT > 0 C) WTOT < 0 20

ACT A car drives up a hill with constant speed. How does the kinetic energy of the car change as it moves up the hill? v A) It increases B) It stays the same C) It decreases 21

Checkpoint A) WTOT = 0 B) WTOT > 0 C) WTOT < 0 A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on the car by all forces as it moves up the hill? v A) WTOT = 0 B) WTOT > 0 C) WTOT < 0 A) since the car moves with constant speed, then its change in kinetic energy is 0, which means that total work done is 0 B) It will take a lot of work to move the truck up the hill-- a positive value. C) The normal force is always parallel to the direction of motion so it does no work but the gravitational force is negative because it acts in a direction opposite the direction of motion. 22

It seems to me that the total work done on an object should equal the change in kinetic energy plus the change in potential energy because if an object gains potential energy some sort of work had to be done on it, but the apple problem in the prelecture stated that it was just the change in kinetic energy. Why is the potential energy not included in this equation? Consider a falling ball. DKE + DPE = 0 We will discuss PE on Wednesday.

Explain how no work is done when you lift an apple to a higher shelf Explain how no work is done when you lift an apple to a higher shelf. there is a force (presumably a hand) going in a parallel direction for x distance (to the higher shelf). using the standard old fashioned W = Fd i don't see how that doesn't work (pun intended) question five in this preflight and question two in the prelecture: I'm getting confused on the wording a little bit. both gain potential energy because they move vertically up, but the question seems to be focused on their kinetic energy without making that distinction in the question itself. Is work only defined by the change in kinetic energy? or is there something in each question that I'm not understanding?

A) Positive B) Negative C) Zero Checkpoint A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. S a The work done on the box by the static frictional force as the truck moves a distance D is: A) Positive B) Negative C) Zero Only 40% got this right… Could we go over the directions of work and go over the friction problem with the boxes? I've tried every number combination available and frankly, the frustration has caused me to break numerous pencils in my dorm room.

A) Positive B) Negative C) Zero Checkpoint A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. S a The work done on the box by the static frictional force as the truck moves a distance D is: A) Positive B) Negative C) Zero

From last time: A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box: S a A B C

A) Positive B) Negative C) Zero Checkpoint F S a D The work done on the box by the static frictional force as the truck moves a distance D is: A) Positive B) Negative C) Zero A) It is positive because the force is in the same direction as the displacement. B) the static friction opposes the motion therefor work is negative C) No work is done since the box is not moving in relation to the truck.

Checkpoint Comments F S a D The work done on the box by the static frictional force as the truck moves a distance D is: The work done on the box must be opposite to the work done by the truck as it accelerates. the frictional force points in the negative direction. Be careful! The friction force and the displacement point in the SAME direction. The net force acted on the box is zero. What force acts to the RIGHT? Total Work done is the change in Kinetic Energy. Since the box does not move its kinetic energy is zero and the work done must also be zero. But the box does move; the box, along with the truck, is even accelerating. The static frictional force and box’s movement are all same direction. The static frictional force works in the positive direction. Ooops! Work does NOT have a direction. That’s important. The work is done in the direction of acceleration, which is the same direction as the truck. Since the velocity of the box is increasing, work is being done. The box isn’t moving. If the box doesn’t move, it will have to fall out the back door because the truck is certainly moving.

Work done by a Spring On the last page we calculated the work done by a constant force when the orientation of the path relative to the force was changing. On this page we will calculate the work done by a spring in one dimension. In this case the orientation of the path relative to the force is simple, but the magnitude of the force changes as we move. If the coordinate system is chosen such that x=0 is the relaxed length of the spring as shown, then the force exerted by the spring on some object attached to its end as function of position is given by Hookes law [F = -kx]. As we move the object between two positions x1 and x2, the force on the object clearly changes. Breaking the movement into tiny steps we see that the work done by the spring along each step will depend on the position : [dW = F(x)*dx = -kxdx]. To find the total work done we need to integrate this expression between x1 and x2 [show]. The mathematics of doing this is straightforward, but its very important to realize what it means conceptually. Evaluating a one dimensional integral of some function F(x) between two points x1 and x2 is really just finding the area under the curve between these points [show] . Evaluating the above integral to find the work done is therefore just finding the area under the force versus position plot between the points x1 and x2. [show] Notice that the formula for the work done by a spring that we just derived depends only on the endpoints of the motion x1 and x2, hence we see that this is a conservative force also. 30

“I am confused about the positive work and negative work and also the positive and negative forces for the spring problems.” Use the formula to get the magnitude of the work Use a picture to get the sign (look at directions) In this example the spring does negative work since F and Dx are in opposite direction. The axes don’t matter.

ACT A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the spring does: A) Positive Work B) Negative Work C) Zero work

ACT A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, your hand does: A) Positive Work B) Negative Work C) Zero work

ACT A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the total work done on the box is: A) Positive B) Negative C) Zero

Work done by gravity

ACT In Case 1 we send an object from the surface of the earth to a height above the earth surface equal to one earth radius. In Case 2 we start the same object a height of one earth radius above the surface of the earth and we send it infinitely far away. In which case is the magnitude of the work done by the Earth’s gravity on the object biggest? A) Case 1 B) Case 2 C) They are the same Case 2 Case 1

ACT Solution Same! Case 1: Case 2: RE Case 2 Case 1 2 RE

Comments and Questions I don’t really get this prelecture at all... So I guess I need help on most of it. Has today’s lecture helped? Does it help to “rewind” the prelecture? Could you go over the work done my gravity again please. I did not get the purpose of the negative sign in the w= - mgy equation. If you make a diagram where y1 is lower than y2, then y2 – y1 is positive or upward and the force mg is negative or downward. The work done by the force of gravity is then negative or W = - mg (y2 – y1). Then – for convenience – we make the equation look “pretty” by choosing to set y1 = 0 and then relabel y2 as simply y so that W = - mg (y – 0) or W = - mgy . I am having a hard time conceptualizing what work actually is. Also, I do not understand why work is only the change in kinetic energy disregarding changes in potential energy. Great observation! Hang on for two more days! We’ll add potential energy tomorrow. Total Work is the change in Kinetic Energy by definition. Part of that total work will be renamed Potential Energy. Just hang on. The sign of the work done is a little confusing. It would be great to have some clarity on that. Also, work done is only due to kinetic energy change or also potential energy. The apple’s case was a little confusing because there is a change in potential energy and so there should be positive work, why zero?? What’s “potential energy”? ; — ) Hang on for two more days. We’ll relabel part of the total work as potential energy tomorrow.