Physics 1501: Lecture 14, Pg 1 Physics 1501: Lecture 14 Today’s Agenda l Midterm graded by next Monday (maybe …) l Homework #5: Due Friday Oct. 11:00 AM l Topics çConservative vs non-conservative forces çConservation of mechanical energy çthe U - F relationship

Physics 1501: Lecture 14, Pg 2 Some Definitions l Conservative Forces - those forces for which the work done does not depend on the path taken, but only the initial and final position. l Potential Energy - describes the amount of work that can potentially be done by one object on another under the influence of a conservative force  W = -  U only differences in potential energy matter.

Physics 1501: Lecture 14, Pg 3 Lecture 14, ACT 1 Work/Energy for Non-Conservative Forces l The air track is once again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v 1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. How much work did friction do on the cart ? 1 meter 30 degrees A) 2.5 J B) 5 J C) 10 J D) –2.5 J E) –5 J F) –10 J

Physics 1501: Lecture 14, Pg 4 Conservation of Energy l If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved. E = K + U  E =  K +  U = W +  U = W + (-W) = 0 using  K = W using  U = -W l Both K and U can change, but E = K + U remains constant. constant!!! E = K + U is constant !!!

Physics 1501: Lecture 14, Pg 5 Example: The simple pendulum. l Suppose we release a mass m from rest a distance h 1 above its lowest possible point. ç What is the maximum speed of the mass and where does this happen ? ç To what height h 2 does it rise on the other side ? v h1h1 h2h2 m

Physics 1501: Lecture 14, Pg 6 Example: The simple pendulum. l Energy is conserved since gravity is a conservative force (E = K + U is constant) l Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice). E = 1 / 2 mv 2 + mgy. v h1h1 h2h2 y y=0

Physics 1501: Lecture 14, Pg 7 Example: The simple pendulum. l E = 1 / 2 mv 2 + mgy. ç Initially, y = h 1 and v = 0, so E = mgh 1. ç Since E = mgh 1 initially, E = mgh 1 always since energy is conserved. y y=0 y=h 1

Physics 1501: Lecture 14, Pg 8 Example: The simple pendulum. l 1 / 2 mv 2 will be maximum at the bottom of the swing. l So at y = 0 1 / 2 mv 2 = mgh 1 v 2 = 2gh 1 v h1h1 y y=h 1 y=0

Physics 1501: Lecture 14, Pg 9 Example: The simple pendulum. l Since E = mgh 1 = 1 / 2 mv 2 + mgy it is clear that the maximum height on the other side will be at y = h 1 =h 2 and v = 0. l The ball returns to its original height. y y=h 1 =h 2 y=0

Physics 1501: Lecture 14, Pg 10 Example: The simple pendulum. l The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1 / 2 mv 2 + mgy = K + U = constant. y Active Figure

Physics 1501: Lecture 14, Pg 11 l What speed will skateboarder reach at bottom of the hill ? R=3 m.. m = 25 kg Initial: K 1 = 0 U 1 = mgR Final: K 2 = 1/2 mv 2 U 2 = 0 Conservation of Total Energy : Lecture 14, Example Skateboard K 1 + U 1 = K 2 + U mgR = 1/2mv v = (2gR) 1/2 v ~ (2 x10m/s 2 x 3m) 1/2 v ~ 8 m/s (~16mph) ! Does NOT depend on the mass !..

Physics 1501: Lecture 14, Pg 12 l What would be the speed if instead the skateboarder jumps to the ground on the other side ?.... R=3 m KINEMATICS: * v = a t * h = 1/2 at 2 2h = v 2 /a v = (2 h a) 1/2 = (2 R g) 1/2 the same magnitude as before ! and independent of mass Lecture 14, Example Skateboard

Physics 1501: Lecture 14, Pg 13 Lecture 14, ACT 2 The Roller Coaster l I have built a Roller Coaster. A motor tugs the cars to the top and then they are let go and are in the hands of gravity. To make the following loop, how high do I have to let the release the car ?? h ? R Car has mass m Hint: from lecture 11 we know that to avoid death, the minimum speed at the top is A) 2RB) 5RC) 5/2 RD) none of the above

Physics 1501: Lecture 14, Pg 14 Non-conservative Forces : l If the work done does not depend on the path taken, the force involved is said to be conservative. l If the work done does depend on the path taken, the force involved is said to be non-conservative. l An example of a non-conservative force is friction: l Pushing a box across the floor, the amount of work that is done by friction depends on the path taken. ç Work done is proportional to the length of the path !

Physics 1501: Lecture 14, Pg 15 Generalized Work Energy Theorem: l Suppose F NET = F C + F NC (sum of conservative and non- conservative forces). l The total work done is: W TOT = W C + W NC The Work Kinetic-Energy theorem says that: W TOT =  K.  W TOT = W C + W NC =  K  W NC =  K - W C But W C = -  U So W NC =  K +  U =  E

Physics 1501: Lecture 14, Pg 16 Lecture 14, ACT 3 Stones and Friction l I throw a stone into the air. While in flight it feels the force of gravity and the frictional force of the air resistance. The time the stone takes to reach the top of its flight path (i.e. go up) is, A) larger than B) equal to C) less than The time it takes to return from the top (i.e. go down).

Physics 1501: Lecture 14, Pg 17 l Let’s now suppose that the surface is not frictionless and the same skateboarder reach the speed of 7.0 m/s at bottom of the hill. What was the work done by friction on the skateboarder ? R=3 m.. m = 25 kg Conservation of Total Energy : Lecture 14, Example Skateboard K 1 + U 1 = K 2 + U 2 W f mgR = 1/2mv W f = 1/2mv 2 - mgR W f = (1/2 x25 kg x (7.0 m/s 2 ) kg x 10m/s 2 3 m) W f = J = J Total mechanical energy decreased by 122 J !.. W f +

Physics 1501: Lecture 14, Pg 18 Conservative Forces and Potential Energy l We have defined potential energy for conservative forces  U = -W l But we also now that W = F x  x l Combining these two,  U = - F x  x l Letting small quantities go to infinitessimals, dU = - F x dx l Or, F x = -dU/dx

Physics 1501: Lecture 14, Pg 19 Examples of the U - F relationship l Remember the spring, çU = (1/2)kx 2 l Do the derivative çF x = - d ( (1/2)kx 2 ) / dx çF x = - 2 (1/2) kx çF x = -kx

Physics 1501: Lecture 14, Pg 20 Examples of the U - F relationship l Remember gravity, çF y = mg l Do the integral çU = -  F dy çU = -  (-mg) dy çU = mgy + C   U = U 2 – U 1 = (mgy 2 + C) – (mgy 1 + C) = mg  y