Welcome back to PHY 183 Meaning of the picture ? PE  KE

3.1 Work 3.2 Energy 3.3 Conservative and nonconservative forces 3.4 Power, energy and momentum conservation 3.5 Linear momentum 3.6 Collisions 3.7 Motion in a gravitational potential CHAPTER 3 WORK AND ENERGY

Part 1 Work Work of constant Force Work done by sum of constant Forces Work done by Variable Force

Review: Work of constant Force F r Work (  W) of a constant force F acting through a displacement  r is: F  r r (N.m=J)  W = F   r = F  r cos(  ) = F r  r (N.m=J)  F rrrr displacement FrFr

Learning check Find work done by tension T and by weigh N Find work done by Normal force N. What is your conclusion from these ?? W T = 0 and W W = ? W N = 0 v N T v

Test: Work done by gravity F  r r. W G =F   r =mg.  r. cos  = -mg  y (remember  y = y f - y i ) W G = -mg  y Depends only on  y ! r Let mass m move on the path  r by gravity only Compute work done ?? j m rrrr gmggmg yy  m

Work done by sum of constant Forces FFFFFr Suppose F NET = F 1 + F 2 + F 3 …+ F n and the displacement is  r. The work done by each force is: F  r F  W 1 = F 1   r W 2 = F 2   r F  …. W n = F n   r W NET = W 1 + W 2 + W 3 …+ W n F  r F  r…+ F  r = F 1   r + F 2   r…+ F n   r F  r W NET = F NET   r Net work done F F TOT rrrr FF1FF1 FF2FF2 FF3FF3 FFnFFn

Test: Work done by gravity r 1 rr n Let mass m move on the path (  r 1 +  r  r n ) by gravity Compute work done on total path: W NET = W 1 + W W n Note: Force = weight= F = mg F  rF  rF  r = F   r 1 + F   r F   r n F  r 1 rr n = F  (  r 1 +  r  r n )  = F   r = F  y Depends only on  y, not on path taken! rr m gmggmg yy j r1r1r1r1 r2r2r2r2 r3r3r3r3 rnrnrnrn

Work done by Variable Force: When the force was constant, we wrote W = F  x –area under F vs. x plot: For variable force, we find the area by integrating: –dW = F(x) dx. F x WgWg xx F(x) x1x1 x2x2 dx

Work of variable force Example: Spring For a spring we know that F x = -kx. F(x) x2x2 x x1x1 -kx relaxed position F = - k x 1 F = - k x 2

Problem Compute work done by the spring W s during a displacement from x 1 to x 2 (Note: the F(x) vs x plot between x 1 and x 2 ). WsWs F(x) x2x2 x x1x1 -kx relaxed position

Solution F(x) x2x2 WsWs x x1x1 -kx

Work by variable force in 3D F l Work dW F of a force F acting through an infinitesimal  r displacement  r is: F  r dW = F.  r l The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: F  r W TOT = F.  rF rrrr 

Part 2 Energy Kinetic energy KE: moving energy Potential energy PE: Tendency for work Total energy: TE=KE + PE

Kinetic Energy Theorem NetWork {Net Work done on object}= changekinetic energy ={change in kinetic energy of object}  W F =  K = 1 / 2 mv / 2 mv 1 2 xxxx F v1v1 v2v2 m W F = F  x

Prove: Kinetic Energy Theorem for a variable Force F F dx dv dx dv v dv v22v22 v12v12 v22v22 v12v12 dv dx v dv dx v(chain rule) dt = =

Test: Falling Objects Three objects of mass m begin at height h with velocity 0. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a pendulum. What is the relationship between their velocities when they have fallen to height 0? v=0 vivi H vpvp vfvf Free Fall Frictionless incline Pendulum (a) (b) (c) (a) V f > V i > V p (b) V f > V p > V i (c) V f = V p = V i

Solution v=0 vivi H vpvp vfvf Free Fall Frictionless incline Pendulum Only gravity will do work: W g = mgH = 1 / 2 mv / 2 mv 1 2 = 1 / 2 mv 2 2 does not depend on path !!

Learning check A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 1 from its relaxed position while momentarily coming to rest. –If the initial speed of the box were doubled and its mass were halved, how far x 2 would the spring compress ? x1x1 (a) (b) (c) (a)  (b) (c)

Solution Again, use the fact that W NET =  K. In this case, W NET = W SPRING = - 1 / 2 kx 2 and  K = - 1 / 2 mv 2 so kx 2 = mv 2 In the case of x 1 So if v 2 = 2v 1 and m 2 = m 1 /2 x1x1 v1v1 m1m1 m1m1

Problem: Spring pulls on mass. A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the relaxed position if it slides without friction? relaxed position stretched position (at rest) d after release back at relaxed position vrvr v m m m m

Step-1 First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring): stretched position (at rest) d relaxed position vrvr m m i

Step-2 Now find the change in kinetic energy of the mass: stretched position (at rest) d relaxed position vrvr m m i

Step-3 Now use work kinetic-energy theorem: W net = W S =  K. stretched position (at rest) d relaxed position vrvr m m i

Step-4 Now suppose there is a coefficient of friction  between the block and the floor f ΔrThe total work done on the block is now the sum of the work done by the spring W S (same as before) and the work done by friction W f. W f = f. Δr = -  mg d stretched position (at rest) d relaxed position vrvr m m i f =  mg r r r r

Step-4 Again use W net = W S + W f =  K W f = -  mg d stretched position (at rest) d relaxed position vrvr m m i f =  mg r r r r

Part 3 Conservative and nonconservative forces

Conservative Forces: conservativeIn general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservative. Gravity is a conservative force: Gravity near the Earth’s surface: A spring produces a conservative force:

Result of conservative Forces We have seen that the work done by a conservative force does not depend on the path taken. W1W1 W2W2 W1W1 W2W2 W 1 = W 2 W NET = W 1 - W 2 = W 1 - W 1 = 0 Therefore the work done in a closed path is zero.

Potential Energy For any conservative force F we can define a potential energy function U in the following way: –The work done by a conservative force is equal and opposite to the change in the potential energy function. This can be written as: Fr W = F. dr = -  U = U 1 – U 2  Fr  U = U 2 - U 1 = -W = - F. dr  r1r1 r2r2 r1r1 r2r2 U2U2 U1U1

Test: Gravitational Potential Energy We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance  y is W g = -mg  y Compute the change in potential energy of this object ??  U = -W g = mg  y yy m W g = -mg  y j

Gravitational Potential Energy So we see that the change in U near the Earth’s surface is:  U = -W g = mg  y = mg(y 2 -y 1 ). arbitrary constant So U = mg y + U 0 where U 0 is an arbitrary constant. Having an arbitrary constant U 0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. y1y1 m W g = -mg  y j y2y2