Physics: Work and Energy

 

Chapter 6 

Work and Energy 

 

 

6.1 Work Done by a Constant Force

W   =  Fs 1 N ⋅  m = 1 joule ( J )

 

 

6.1 Work Done by a Constant Force

 

 

6.1 Work Done by a Constant Force

cos 0 = 1 

  cos θ  ) s W  = ( F   s

 

cos 90 = 0 

cos180 = −1 

 

6.1 Work Done by a Constant Force

Example 1 Pulling a Suitcase-on-W Suitcase-on-Wheels heels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m. W  = ( F    cos θ  ) s = [ ( 45.0 N ) cos 50.0 ]( 75.0 m )  

=

2170 J  



 

6.1 Work Done by a Constant Force

  W  = ( F  cos 0) s  = Fs

W  = ( F  cos  180) s =  − Fs

 

 

6.1 Work Done by a Constant Force

Example 3 Accelerating a Crate The truck is accelerating at a rate of +.50 m!s ". The mass of the crate is "0-kg and it does not slip. The magnitude of  the displacement is #5 m. $hat is the total work done on the crate %& all of the forces acting on it'

 

 

6.1 Work Done by a Constant Force

The angle %etween the displacement and the normal force is (0 degrees. The angle %etween the displacement and the weight is also (0 degrees.

W  = ( F  cos   90  ) s  = 0

 

 

6.1 Work Done by a Constant Force

The angle %etween the displacement and the friction force is 0 degrees.

 f   s = ma = (120 kg ) (1.5 m s  

2

) = 180 N

W  = [ (180 N ) cos 0]( 65  m ) = 1.2 × 10 4 J

 

 

6.2 The Work-Energy Theorem and !net!c Energy 

)onsider a constant net e*ternal force acting on an o%ect. The o%ect is displaced a distance s, in the same direction as the net force.

∑ F   s The work is simpl&

 

W  =  ( ∑   F ) s  = ( ma ) s

 

6.2 The Work-Energy Theorem and !net!c Energy 

W  = m( as )

=

  m

1 2

( v 

2  f  

2 o

−v

) =   mv 1 2

2   f  

−

1 2

v f  2 = vo2 + 2( ax )

( ax ) =  12  ( v f  2 −  vo2 ) FNT/N /F NT) N123 The kinetic energ&  of and o%ect with mass m and speed v  is  is gien %& 1 2

 

KE =  mv

2

2 o

mv

 

6.2 The Work-Energy Theorem and !net!c Energy 

T $/1-N123 T/16 $hen a net e*ternal force does work on and o%ect, the kinetic energ& of the o%ect changes according to

f  −   o W  = KE   KE  

=

1 2

2

mvf 

−

1 2

2

mvo

 

6.2 The Work-Energy Theorem and !net!c Energy 

Example " Deep Space 1 The mass of the space pro%e is 474-kg and its initial elocit& is "75 m!s. f the 5#.0-mN 5#.0-mN force acts on the pro%e pro%e through a displacement displace ment of ".4"0(m, what is its final speed'

 

 

6.2 The Work-Energy Theorem and !net!c Energy 

2 f 

2 o

W = mv   − mv 1 2

W = [ ( ∑ F ) cos   θ ] s  s

 

1 2

 

6.2 The Work-Energy Theorem and !net!c Energy 

[ ( ∑ F) cos   θ ] s = ( 5.60 ×10

1 2

 2 f 

1 2

2 o

 mv − mv

 N ) cos 0 ( 2.42 ×   109 m ) =  12 ( 474 kg ) vf 2 − 12 ( 474 kg ) ( 275 m s )

-2



v f   = 805 m s

 

2

 

6.2 The Work-Energy Theorem and !net!c Energy 

  25 −  f  k  n this case the net force is ∑ F  = mg sin  

 



 

6.2 The Work-Energy Theorem and !net!c Energy 

Concept#al Example 6 Work and Kinetic Energy  8 satellite is moing a%out a%out the earth in a circular or%it and an elliptical or%it. For these two or%its, determine whether  the kinetic energ& of the satellite changes during the motion.

 

 

6.3 $ra%!tat!onal &otent!al Energy 

W  = ( F    cos θ  ) s  s W g!"i#$ =  mg ( ho −  h f   )

 

 

6.3 $ra%!tat!onal &otent!al Energy 

W g!"i#$ =  mg ( ho −  h f   )

 

 

6.3 $ra%!tat!onal &otent!al Energy 

Example ' A Gymnast on a Trampoline The g&mnast leaes the trampoline at an initial height of ."0 m and reaches a ma*imum height of 4.90 m %efore falling %ack down. $hat was the initial speed of the g&mnast'

 

 

6.3 $ra%!tat!onal &otent!al Energy 

2 f 

2 o

W = mv   − mv 1 2

1 2

W g!"i#$ =  mg ( ho −  h f   )

1 2

2 o

mg ( ho −  h f   )  = − mv

vo = −  2  g ( ho − h f   )

vo = − 2( 9  .80 m s 2 )(1.20 m − 4.80 m ) = 8.40 m s

 

6.3 $ra%!tat!onal &otent!al Energy 

W g!"i#$ =  mgho −  mgh f  

FNT/N /F 218:T8T/N8;

E =   E f 

o

   

6.* The Conser%at!on o+ ,echan!cal Energy 

T