Physics 211 - PowerPoint

Document Sample
Physics 211 - PowerPoint Powered By Docstoc
					                        Stuff you asked about:                         5x pudding
Talk more about work.
go over boxes being pulled on a table by a string connected to a box
hanging off the table
How much of the calculus do we need to know?
Why did you change from fading in each equation and circling the important
bits to just throwing all the integrations and whatnot onto one eye-bending
page?

all of them, especially the ball on the curved slope problem

I'm incredibly unfamiliar with vector calculus, and I'd like you to clarify how to
combine vectors with integrals properly. Seriously. It's making me rethink
how I've been doing physics in high school and kind of scaring me.

The physics of massive amounts of bunnies in a very small box that is
moving across a pile of frictionless bunnies with a constant acceleration. I
don't know if there is an equation for that...
Please put this on the lecture slides in class tomorrow. It's my dream to be
on it.
                                                             Physics 211 Lecture 7, Slide 1
  Physics 211
     Lecture 7
 Today's Concepts:
Work & Kinetic Energy




                   Physics 211 Lecture 7, Slide 2
The Dot Product




                  Physics 211 Lecture 7, Slide 3
            Work Kinetic Energy Theorem

The work done by force F as it acts on an object that moves
between positions r1 and r2 is equal to the change in the
object’s kinetic energy:


                   W  K
             r2
                                           1
      W   F  dl                      K  mv 2

             r1
                                           2

                                            Physics 211 Lecture 7, Slide 4
If the force is constant and the directions aren’t changing
then this is very simple to evaluate:




                        F       d
               Car



                      r2

             W   F  dl  F  d
                       r1

                     In this case    Fd   since cos(0)=1

                                              Physics 211 Lecture 7, Slide 5
                             Act
A lighter car and a heavier truck, each initially at rest, are
pushed with the same constant force F. After both vehicles
travel a distance d, which of the following statements is
true? (Ignore friction)


                         F    d
                   Car


                         F    d
           Truck



    A) The will have the same velocity
    B) They will have the same kinetic energy
    C) They will have the same momentum
                                             Physics 211 Lecture 7, Slide 6
  r2
                          Derivation – not so important
   F dl
  r
    F dl      K         Concept – very important
   1




                                              dl
                                             q

A force pushing over some distance
will change the kinetic energy                        F
                                           Physics 211 Lecture 7, Slide 7
Work done by gravity near the Earth’s surface




                         mg




                                Physics 211 Lecture 7, Slide 8
  Work done by gravity near the Earth’s surface


      WTOT  W1  W2  ...  WN                                    dlN
 mg  dl1  mg  dl2  ...  mg  dlN

                                         dl1
                                          dl2
                                                                      dl1
                                   mg                        dy1
                                                                    dx1

                                                               mg



                                                Physics 211 Lecture 7, Slide 9
  Work done by gravity near the Earth’s surface


      WTOT  W1  W2  ...  WN                                    dlN
 mg  dl1  mg  dl2  ...  mg  dlN                                    y
 mgdy1  mgdy2 ...  mgdyN             dl1
 mg y                                  dl2
                                   mg



                Wg  mg y
                                                Physics 211 Lecture 7, Slide 10
              Work Kinetic Energy Theorem
 If there are several forces acting then W is the work done by the
 net (total) force:

                   WNET  K
         r2

WNET   ( F1  F2  ...)  dl
         r1

         r2           r2

         F1  dl   F2  dl1  ...
         r1           r1

                             You can just add up the
        W1  W2  ...       work done by each force
                                              Physics 211 Lecture 7, Slide 11
                          Preflight
Three objects having the same mass begin at the same
height, and all move down the same vertical distance H. One
falls straight down, one slides down a frictionless inclined
plane, and one swings on the end of a string.
In which case does the object have the biggest net work
done on it by all forces during its motion?




                H


    Free Fall         Frictionless incline       String

 A) Free Fall       B) Incline    C) String D) All the same
                                   Act
Three objects having the same mass begin at the same
height, and all move down the same vertical distance H. One
falls straight down, one slides down a frictionless inclined
plane, and one swings on the end of a string.
What is the relationship between their velocities when they
reach the bottom?




                H


    Free Fall             Frictionless incline           String

        A) Vf > Vi > Vp        B) Vf > Vp > Vi   C) Vf = Vp = Vi
                    Preflight/Act



               H


  Free Fall            Frictionless incline             String

     A) Vf > Vi > Vp      B) Vf > Vp > Vi     C) Vf = Vp = Vi


Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22

                            v f  v i  v p  2 gH
                              Preflight/Act
WNET  K
A car drives up a hill with constant speed. Which
statement best describes the total work WTOT done on
the car by all forces as it moves up the hill?

 A) WTOT = 0
 B) WTOT > 0
 C) WTOT < 0


A) velocity is not changing, and therefore the work (which equals change in
kinetic energy) is zero.
B) The net force and the distance are in the same direction.
C) The car is moving against gravity so work is negative.

                                                      Physics 211 Lecture 7, Slide 15
                   From Monday
A box sits on the horizontal bed of a moving truck. Static
friction between the box and the truck keeps the box from
sliding around as the truck drives.


                                        S
          a

If the truck moves with constant accelerating to the left as
shown, which of the following diagrams best describes the
static frictional force acting on the box:



                    A        B          C

                                             Physics 211 Lecture 7, Slide 16
                     Preflight/Act
A box sits on the horizontal bed of a moving truck which
his accelerating to the left as shown. Static friction
between the box and the truck keeps the box from
sliding.

                              F           S
          a

              D
 The work done on the box by the static frictional force as the
 truck moves a distance D is:

       A) Positive          B) Negative            C) Zero


                                               Physics 211 Lecture 7, Slide 17
          Work done by a spring
                          0       x1



                   FS

                                                            x



                       FS = -kx

    x2            x1                   x1
                                1 2    1 2
W   F ( x)  dx  k  xdx   kx   kx1
    x1                 0        2   0  2
                                       Physics 211 Lecture 7, Slide 18
                             Act
A box sliding on a horizontal frictionless surface runs into a
fixed spring, compressing it a distance x1 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 x2 would the spring compress ?
       A)   x2  x1   B)   x2  2 x1   C)   x2  2 x1


                                 x1

                                                              1 2
                                                        WS   kx1
                                                              2
              Act Solution
     v
                            x1




                1 2             1 2
   Wspring     kx1      K   mv
                2               2
       m           If v is doubled and m is halved:
x1  v
       k                   m
                   x2  2v     2 x1
                           2k
                                    Physics 211 Lecture 7, Slide 20
          Work done by gravity in general




     r2             r2
                                                    1 1
                                        r
                         GM e m   GM e m 2
W   F (r )  dr        2              GM e m   
    r1                r1
                          r         r r1             r2 r1 

                                            Physics 211 Lecture 7, Slide 21
                           Act
In Case 1 we send an object from the surface of the earth to
a height above the earth surface equal to one earth radius.

In Case 2 we start the same object a height of one earth
radius above the surface of the earth and we send it infinitely
far away.

   In which case is the magnitude of the work done by the
   Earth’s gravity on the object biggest?

     A) Case 1        B) Case 2        C) They are the same
                   Act Solution
                    1     1    GM e m
Case 1: W  GM e m         
                    2 RE RE     2 RE
                                                         Same!
                       1    1      GM e m
Case 2:     W  GM e m         
                         2 RE      2 RE




RE

          2RE                          Physics 211 Lecture 7, Slide 23

				
DOCUMENT INFO