康涅狄格州大学：《普通物理》课程PPT教学课件（英文版）Lecture 13 Work-kinetic energy theorem

Physics 121, Sections 9, 10, 11, and 12 Lecture 13 Today's Topics Homework 6: due after midterm week Chapter 6: Work and Energy Definition of work Work-kinetic energy theorem Potential energy Non-conservative forces Generalized work-kinetic energy theorem Examples Physics 121: Lecture 13, Pg 1

Physics 121: Lecture 13, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 13 Today’s Topics: Homework 6: due after midterm week Chapter 6: Work and Energy Definition of work Work-kinetic energy theorem Potential energy Non-conservative forces Generalized work-kinetic energy theorem Examples

Definition of work Ingredients: Force(F), displacement (Ar) Work W. of a constant force F acting through a displacement A r is F W=F△rcos0 0/△r Physics 121: Lecture 13, Pg 2

Physics 121: Lecture 13, Pg 2 Definition of Work: Ingredients: Force ( F ), displacement (  r ) Work, W, of a constant force F acting through a displacement  r is: W = F  r cos   F  r Fr

Units. Force x Distance= Work Newton x Meter Joule [M[L/2[][ML]2/[2 mks gs other N-m(Joule) Dyne-cm(erg) BTU 1054J 107J calorie = 4.184 J foot-Ib =1.356 J e∨ 1.6×10-19J Physics 121: Lecture 13, Pg 3

Physics 121: Lecture 13, Pg 3 Units: N-m (Joule) Dyne-cm (erg) = 10-7 J BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J mks cgs other Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2

Work Kinetic-Energy Theorem INet Work done on object] Change in kinetic energy of object W=△K K-K mv mV Physics 121: Lecture 13, Pg 4

Physics 121: Lecture 13, Pg 4 Work Kinetic-Energy Theorem: {Net Work done on object} = {change in kinetic energy of object} Wnet = K = K2 − K1 2 1 2 2 mv 2 1 mv 2 1 = −

Work done by Variable Force:(1D) When the force was constant we wrote W= FAX area under F vs x plot △X F(x) ∠X Physics 121: Lecture 13, Pg 5

Physics 121: Lecture 13, Pg 5 Work done by Variable Force: (1D) When the force was constant, we wrote W = Fx area under F vs x plot: F x Wg x F(x) x1 x2 x

Springs A very common problem with a variable force is a spring L F △x In this spring, the force gets greater as the spring is further compressed Hooks Law k△x Active Figure AX is the amount the spring is stretched or compressed from it resting position Physics 121: Lecture 13, Pg 6

Physics 121: Lecture 13, Pg 6 Springs A very common problem with a variable force is a spring. In this spring, the force gets greater as the spring is further compressed. Hook’s Law, FS = - k x x is the amount the spring is stretched or compressed from it resting position. F x Active Figure

1-D Variable Force Example: Spring For a spring we know that Fx=-kx(Hook's law) F(×) X relaxed position WA F=-k F=-k Physics 121: Lecture 13, Pg 7

Physics 121: Lecture 13, Pg 7 1-D Variable Force Example: Spring For a spring we know that Fx = -kx (Hook’s law). F(x) x2 x x1 -kx relaxed position F = - k x1 F = - k x2

Spring… The work done by the spring Ws during a displacement from X, to x2 is the area under the F(x)vs x plot between x and x F(×) 1 X relaxed position WA Physics 121: Lecture 13, Pg 8

Physics 121: Lecture 13, Pg 8 Spring... The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between x1 and x2 . Ws F(x) x2 x x1 -kx relaxed position

Spring The work done by the spring Ws during a displacement from x, to x, is the area under the F(x)vs x plot between X, and x2 F(×) Ws=-1/2[(kx2)(x2)-(kx1)(x1)] kX Physics 121: Lecture 13, Pg 9

Physics 121: Lecture 13, Pg 9 Spring... The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between x1 and x2 . x1 x2 F(x) x Ws kx1 kx2 -kx Ws Ws = - 1/2 [ ( kx2 ) (x2 ) - (kx1 ) (x1 ) ]

Lecture 13. AcT 1 Work Energy a box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 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 would the spring compress? (b)x=√2x(c)x=2X Physics 121: Lecture 13, Pg 10

Physics 121: Lecture 13, Pg 10 Lecture 13, ACT 1 Work & Energy A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 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’ would the spring compress ? x (a) x' = x (b) x' = 2 x (c) x' = 2 x