Phys 207. Announcements. Hwk 6 is posted online; submission deadline = April 4 Exam 2 on Friday, April 8th. Today s Agenda

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1 Phys 07 Announceents Hwk 6 is posted online; subission deadline = April 4 Ea on Friday, April 8th Today s Agenda Review of Work & Energy (Chapter 7) Work of ultiple constant forces Work done by gravity near the Earth s surface Eaples: pendulu, inclined plane, free fall Work done by variable force Spring Proble involving spring & friction Review: Constant Force Work, W, of a constant force F acting through a displaceent is: W = Fi = F cos(θ) = F r F θ F r displaceent Page

2 Review: Work/Kinetic Energy Theore: {Net Work done on object} = {change in kinetic energy of object} W F = K = / v - / v v v F W F = F 3 A siple application: Work done by gravity on a falling object What is the speed of an object after falling a distance H, assuing it starts at rest? W g = Fi = g cos(0) = gh v 0 = 0 W g = gh H g j Work/Kinetic Energy Theore: W g = gh = / v v = gh v 4 Page

3 What about ultiple forces? Suppose F NET = F + F and the displaceent is. The work done by each force is: F NET W = F i W = F i F W TOT = W + W = F i + F i = (F + F )i F W TOT = F TOT i It s the total force that atters!! 5 Coents: Tie interval not relevant Run up the stairs quickly or slowly...sae W Since W = Fi No work is done if: F = 0 or = 0 or θ = 90 o 6 Page 3

4 Coents... W = Fi No work done if θ = 90 o. No work done by T. T v No work done by N. N v 7 Lecture 5, Act Work & Energy An inclined plane is accelerating with constant acceleration a. A bo resting on the plane is held in place by static friction. How any forces are doing work on the block? a (a) (b) (c) 3 8 Page 4

5 Lecture 5, Act Solution First, draw all the forces in the syste: F S a g N 9 Lecture 5, Act Solution Recall that W = Fi so only forces that have a coponent along the direction of the displaceent are doing work. F S a g N The answer is (b). 0 Page 5

6 Work done by gravity: W g = Fi = g cos θ = -g y W g = -g y Depends only on y! θ g y j Work done by gravity... W NET = W + W W n = Fi + Fi Fi n = Fi ( n ) = Fi = F y W g = -g y y 3 g j Depends only on y, not on path taken! n Page 6

7 Lecture 5, Act Falling Objects Three objects of ass 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 pendulu. What is the relationship between their velocities when they have fallen to height 0? v=0 v=0 v=0 H v f v i v p Free Fall Frictionless incline Pendulu (a) V f > V i > V p (b) V f > V p > V i (c) V f = V p = V i 3 v = 0 Lecture 5, Act Solution v = 0 v = 0 H v f v i v p Free Fall Frictionless incline Pendulu Only gravity will do work: W g = gh = / v - / v = / v v = v = v gh does not depend on path!! f i p = 4 Page 7

8 Lifting a book with your hand: What is the total work done on the book?? First calculate the work done by gravity: W g = gi = -g Now find the work done by the hand: F HAND v = const a = 0 W HAND = F HAND i = F HAND g 5 Eaple: Lifting a book... W g W HAND W NET = -g = F HAND = W HAND + W g = F HAND - g = (F HAND - g) F HAND g v = const a = 0 = 0 since K = 0 (v = const) So W TOT = 0!! 6 Page 8

9 Eaple: Lifting a book... Work/Kinetic Energy Theore says: W = K {Net Work done on object} = {change in kinetic energy of object} In this case, v is constant so K = 0 and so W ust be 0, as we found. F HAND v = const a = 0 g 7 Work done by Variable Force: (D) When the force was constant, we wrote W = F area under F vs. plot: F W g For variable force, we find the area by integrating: dw = F() d. F() W = F( )d d 8 Page 9

10 Work/Kinetic Energy Theore for a Variable Force W = F d = dv dt v = v v dv d d d F = a = dv dt dv d dv dv v dt = = dt d d (chain rule) v = v dv v = ( v ) = = KE v v v 9 -D Variable Force Eaple: Spring For a spring we know that F = -k. F() relaed position -k F= -k F= -k 0 Page 0

11 Spring... The work done by the spring W s during a displaceent fro to is the area under the F() vs plot between and. F() relaed position -k W s Spring... The work done by the spring W s during a displaceent fro to is the area under the F() vs plot between and. F() W s -k W s = F( ) d = ( k) d = k W s = k ( ) Page

12 Lecture 5, Act 3 Work & Energy A bo sliding on a horizontal frictionless surface runs into a fied spring, copressing it a distance fro its relaed position while oentarily coing to rest. If the initial speed of the bo were doubled and its ass were halved, how far would the spring copress? = (a) (b) = (c) = 3 Lecture 5, Act 3 Solution Again, use the fact that W NET = K. In this case, W NET =W SPRING = - / k and K = - / v so k = v In the case of = v k v 4 Page

13 = v k Lecture 5, Act 3 Solution So if v = v and = / = v = v k k = v 5 Proble: Spring pulls on ass. A spring (constant k) is stretched a distance d, and a ass is hooked to its end. The ass is released (fro rest). What is the speed of the ass when it returns to the relaed position if it slides without friction? relaed position stretched position (at rest) d after release v back at relaed position v r 6 Page 3

14 Proble: Spring pulls on ass. First find the net work done on the ass during the otion fro = d to = 0 (only due to the spring): W = k = s ( ) = k( 0 d ) kd stretched position (at rest) v r d relaed position i 7 Proble: Spring pulls on ass. Now find the change in kinetic energy of the ass: K = v v = v r stretched position (at rest) v r d relaed position i 8 Page 4

15 Proble: Spring pulls on ass. Now use work kinetic-energy theore: W net = W S = K. kd = v r v r = d k stretched position (at rest) d relaed position v r i 9 Proble: Spring pulls on ass. Now suppose there is a coefficient of friction µ between the block and the floor The total work done on the block is now the su of the work done by the spring W S (sae as before) and the work done by friction W f. W f = f. = - µg d stretched position (at rest) d f= µg relaed position v r i 30 Page 5

16 Proble: Spring pulls on ass. Again use W net = W S + W f = K W f = -µg d WS = kd K = v r kd µ gd = v r k vr = d µ gd stretched position (at rest) v r d f= µg relaed position i 3 Page 6