Lecture09

Physics 101: Lecture 9 Work and Kinetic Energy ,[object Object],Exam II

Energy ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],12

Energy is Conserved ,[object Object],[object Object],[object Object],[object Object],[object Object],10

Work: Energy Transfer due to Force ,[object Object],[object Object],[object Object],[object Object],[object Object],15 W = F dcos(  ) T a mg T b mg T b

Work by Constant Force ,[object Object],[object Object],W F < 0: 90<  < 180 : cos(  ) < 0 W F = 0:  =90 : cos(  ) =0 W F < 0: 90<  < 180 : cos(  ) < 0 W F > 0: 0<  < 90 : cos(  ) > 0 18 A) W>0 B) W=0 C) W<0 1) 2) 3) 4) Note Change in r! F   r F  r F  r F  r F  r F   r F 

ACTS: Ball Toss ,[object Object],[object Object],[object Object],[object Object],[object Object],20

Work by Constant Force ,[object Object],30 50 N W = F  x cos     m ) cos (30) = 217 Joules 21 T mg N f

Where did the energy go? ,[object Object],[object Object],[object Object],W = F  r cos   30 x 5 cos(90)  X-Direction:  F = ma T cos(30) – f = 0 f = T cos(30) W = F  r cos   50 cos(30) x 5 cos(180)  Joules 25 mg 90  r T mg N f f  r 180

Preflight 1 ,[object Object],[object Object],T V “ Normal force is perpendicular to the direction of the car, so it does not contribute” 28 W F N correct

Preflight 2 ,[object Object],[object Object],“ The gravitational force is acting against the car going up the hill which makes it negative” 30 W T F N V correct

Preflight 3 ,[object Object],[object Object],“ The work done by the tow rope would be positive because the distance and force applied by the tow rope are both in the positive direction” 32 W T F N V correct

Kinetic Energy: Motion ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],35

Preflight 4 ,[object Object],[object Object],The net acceleration is 0 which means the sum of the forces is 0. Thus, work done is 0.. 37 W T F N V correct

Example: Block w/ friction ,[object Object],5 m/s Y direction:  F=ma N-mg = 0 N = mg Work W N = 0 W mg = 0 W f = f  x cos(180) = -  mg  x W =  K -  mg  x = ½ m (v f 2 – v 0 2 ) -  g  x = ½ (0 – v 0 2 )  g  x = ½ v 0 2  x = ½ v 0 2 /  g = 3.1 meters 44 mg N f x y

Falling Ball Example ,[object Object],Only force/work done by gravity  W =  KE W g = ½ m (v f 2 – v i 2 ) F g h cos(0) = ½m v f 2 mgh = ½m v f 2 V f = sqrt( 2 g h ) = 10 m/s mg 47

Work by Variable Force ,[object Object],[object Object],[object Object],[object Object],Force Distance Work Work 49 Force Distance F=kx

Summary ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],50

Lecture09

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