Stuff you asked about: I thought I understood the concepts in the pre-lecture until they went over the three examples. I felt that it moved too fast especially when the video was showing manipulations of equations. I needed so much help with this. Please explain all of the checkpoint and lecture questions. Also, give more examples!!! I think work and energy is my favorite topic. I think i know the answers, but don't know why and how to explain why it is right. Well, the day I've been dreading has come. We have finished everything I've previously known about physics and the rest of the semester is new. Time to start learning. We should discuss calc 2 because it is much more difficult than this. This isnt that confusing to me Many of the concepts of work are very confusing. I hope to go over this in detail tomorrow 1. This stuff is super confusing. Discuss all of it. 2. I think everyone will be glad to know the second I read about centrifugal force in my MCB lab manual, I dropped the class. DON'T THEY KNOW PHYSICS?!?! That was a lot to cover in one prelecture. How is dot product different from vector sum?

“It seems like work was defined differently to me in high school than it is here. Then again, I didn't pay much attention in high school” You are old enough & smart enough to see where the formulas come from now.

You aren't supposed to know everything after the prelecture “I really just don't understand work. Please try to cover this in a very understandable way. The pre-lecture just didn't do it for me! :)” You aren't supposed to know everything after the prelecture

Today’s Concepts: Work & Kinetic Energy Physics 211 Lecture 7 Today’s Concepts: Work & Kinetic Energy

Work-Kinetic Energy Theorem “Integrals are a little scary, but the concepts themselves don't seem to difficult.!” The integral IS the concept !!

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: TOT TOT NET “Doesn't work-kinetic energy theorem have anything to do with potential energy?."

Clicker Question Does the work-kinetic energy theorem have anything to do with potential energy? A) YES B) NO

“Direction and magnitude of the total work done is confusing..” The Dot Product “Direction and magnitude of the total work done is confusing..” “the vector dot thing. what is the point of it? Will we have to use this in problems.”

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 = Fd In this case since cos(0)=1 This is probably what you remember from High School.

W= Fd DK= W Clicker Question 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 W= Fd same car F d DK= W Same too 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 “May we please discuss the concepts of gravitational work?” 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

Fg dr dr rdq

Close to the Earth’s surface: r1 ~ r2 ~ Re: mg -Dy So: Wg = -mgDy

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

Clicker Question 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 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 Less that 40% got this right… 20

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

Clicker Question A car drives up a hill with constant speed. The acceleration of the car: A) Points up the hill B) Points down the hill C) Is zero v 22

A car drives up a hill with constant speed. The net force on the car: Clicker Question A car drives up a hill with constant speed. The net force on the car: A) Points up the hill B) Points down the hill C) Is zero v 23

We know two expressions that involve WTOT Reminder We know two expressions that involve WTOT TOT TOT NET

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 TOT A) The car moves upward, in the positive direction, therefore the force that caused it to move by a positive distance did positive work. B) change in kinetic energy is zero, total work done is zero. C) the work is being done by gravity, which is in the negative direction.. 25

“Question 1 in the prelecture is wrong “Question 1 in the prelecture is wrong. It asks for the work done and not the change in kinetic energy. Since the Apple is raised, there is work from the change in gravitational potential energy.” Hand does positive work. Gravity does negative work. The sum of these is zero since Wtot = Whand + Wgravity = DK = 0

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 Less that 40% got this right… “How friction is related to work, and which direction work goes in various cases”

From Last Lecture 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 accelerating 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) Zero C) Negative 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) Zero C) Negative A) since the static frictional force is acting in the same direction as the motion, i believe the work would be positive. B) Static means the box isn't moving, so the displacement is 0. C) Friction always does negative work.

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

“Can you clarify the directions of positive or negativity for springs when they are compressing and uncompressing?” 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.