1
00:00:03,530 --> 00:00:06,750
Let's consider an example
of motion, in which we want

2
00:00:06,750 --> 00:00:08,760
to use our energy concepts.

3
00:00:08,760 --> 00:00:14,850
So suppose we have a ramp
which is circular, radius r.

4
00:00:14,850 --> 00:00:17,030
And we have an object here.

5
00:00:17,030 --> 00:00:21,370
And now, here's a surface
a certain distance,

6
00:00:21,370 --> 00:00:28,530
d, that has friction, where
the coefficients of friction

7
00:00:28,530 --> 00:00:30,470
is non-uniform.

8
00:00:30,470 --> 00:00:35,045
So we'll write it mu-0 And
we'll write it as mu-1x.

9
00:00:35,045 --> 00:00:37,560
We'll take x equals zero.

10
00:00:37,560 --> 00:00:46,680
And then here, there's
a spring and a wall.

11
00:00:46,680 --> 00:00:49,830
And here there is no friction.

12
00:00:49,830 --> 00:00:56,130
Now as we drop, assuming we
have enough height here, h,

13
00:00:56,130 --> 00:00:58,590
so that this block will
slide across the friction,

14
00:00:58,590 --> 00:01:02,250
make it all the way across and
start compressing its spring.

15
00:01:02,250 --> 00:01:06,360
What I'd like to
find is how much

16
00:01:06,360 --> 00:01:07,880
the spring has been compressed.

17
00:01:15,220 --> 00:01:17,440
OK, now how do we analyze this?

18
00:01:17,440 --> 00:01:21,280
Well the key is to use
our energy principle,

19
00:01:21,280 --> 00:01:26,710
where we have w-external
equals the change

20
00:01:26,710 --> 00:01:29,920
in mechanical energy.

21
00:01:29,920 --> 00:01:32,289
And the key is to,
what we're going

22
00:01:32,289 --> 00:01:34,060
to do is the tool that
we're going to use

23
00:01:34,060 --> 00:01:41,979
is what we call energy
diagrams for the initial.

24
00:01:41,979 --> 00:01:43,780
So what does this mean?

25
00:01:43,780 --> 00:01:47,600
We want to choose, we
want to first identify

26
00:01:47,600 --> 00:01:53,360
the initial and final states
that we're referring to.

27
00:01:53,360 --> 00:01:59,229
So, in our picture, here
is our initial state.

28
00:01:59,229 --> 00:02:03,760
And I'll draw the final state
in when the spring has been

29
00:02:03,760 --> 00:02:08,470
compressed a distance, x-final.

30
00:02:08,470 --> 00:02:09,910
So I drew it on my diagram.

31
00:02:12,640 --> 00:02:14,920
So we have initial
and final states.

32
00:02:14,920 --> 00:02:19,720
And now what we
want to do is choose

33
00:02:19,720 --> 00:02:24,910
reference points,
zero-points, zero-point

34
00:02:24,910 --> 00:02:27,100
for each potential function.

35
00:02:31,040 --> 00:02:33,720
And show that on our diagram.

36
00:02:33,720 --> 00:02:37,670
So for the potential
energy of gravity, here,

37
00:02:37,670 --> 00:02:44,390
if we chose this to be
y, then this y equals 0.

38
00:02:44,390 --> 00:02:48,740
And u at y equals 0 is 0.

39
00:02:48,740 --> 00:02:51,829
And we'll call that the
zero point for gravity.

40
00:02:51,829 --> 00:02:54,590
And the zero point
for the spring

41
00:02:54,590 --> 00:03:00,530
is when this x, now I'm going
to call here x equals 0.

42
00:03:00,530 --> 00:03:08,240
So, let's call this a variable,
I can call it anything I want.

43
00:03:08,240 --> 00:03:11,900
I'll call it u-final.

44
00:03:11,900 --> 00:03:17,300
And this is where u is zero.

45
00:03:17,300 --> 00:03:22,220
So u-final just measures
the stretch of the spring.

46
00:03:22,220 --> 00:03:26,300
And so at y equals 0 is 0.

47
00:03:26,300 --> 00:03:31,610
And u spring little
u equal 0, 0.

48
00:03:31,610 --> 00:03:35,280
So now I can
identify my energies.

49
00:03:35,280 --> 00:03:38,030
So let's talk about
the initial energy

50
00:03:38,030 --> 00:03:41,090
is all gravitational potential.

51
00:03:41,090 --> 00:03:43,190
That's mgy.

52
00:03:43,190 --> 00:03:46,550
It's starting at rest.

53
00:03:46,550 --> 00:03:51,800
And e-final well, here,
this is the distance

54
00:03:51,800 --> 00:03:53,990
where it comes to rest, also.

55
00:03:53,990 --> 00:03:56,280
So there's no final
kinetic energy.

56
00:03:56,280 --> 00:03:58,470
There's no gravitational
potential energy,

57
00:03:58,470 --> 00:04:00,560
because we're on the
surface at y equals 0.

58
00:04:00,560 --> 00:04:02,330
But our spring has
been compressed

59
00:04:02,330 --> 00:04:05,134
by 1/2 k little-u-final square.

60
00:04:08,180 --> 00:04:12,580
Now, in terms of
our external work

61
00:04:12,580 --> 00:04:16,010
equals the change in
mechanical energy,

62
00:04:16,010 --> 00:04:18,860
we have now identified
the right-hand side

63
00:04:18,860 --> 00:04:21,769
using these tools of
the energy diagrams.

64
00:04:21,769 --> 00:04:27,500
And I can write my description
u-final squared, minus mgy.

65
00:04:27,500 --> 00:04:30,710
Now what I have to do is
think about the friction

66
00:04:30,710 --> 00:04:36,380
force, f kinetic friction,
as the object moves.

67
00:04:36,380 --> 00:04:39,140
This friction force
is non-uniform.

68
00:04:39,140 --> 00:04:47,780
If we were to draw n, and mg,
then our friction force here

69
00:04:47,780 --> 00:04:53,000
is equal to the integral minus
the integral from x equals 0,

70
00:04:53,000 --> 00:04:55,730
to x equals d.

71
00:04:55,730 --> 00:05:00,540
Equals 0 to x equal d,
of the friction force,

72
00:05:00,540 --> 00:05:04,890
which is equal to the
coefficient of friction, ukndx.

73
00:05:08,720 --> 00:05:16,130
Now that 0 to d, notice that
our coefficient of friction

74
00:05:16,130 --> 00:05:17,380
is varying.

75
00:05:17,380 --> 00:05:19,910
And I chose it intentionally
to show you that friction

76
00:05:19,910 --> 00:05:21,590
is really an integral.

77
00:05:21,590 --> 00:05:26,300
So we have mu-0 plus mu-1 xmgdx.

78
00:05:30,710 --> 00:05:33,470
Let's put our integration
variable in there.

79
00:05:33,470 --> 00:05:35,310
And these are just two
separate integrals.

80
00:05:35,310 --> 00:05:37,159
The first one is easy.

81
00:05:37,159 --> 00:05:40,430
It's minus mu-0 mgd.

82
00:05:40,430 --> 00:05:44,870
And the second one is -mu-1 mg.

83
00:05:44,870 --> 00:05:47,900
And we're just integrating
x-prime, dx prime

84
00:05:47,900 --> 00:05:49,580
between 0 and d.

85
00:05:49,580 --> 00:05:53,870
So that's simply
d-squared over 2.

86
00:05:53,870 --> 00:05:59,420
And that's equal to
ku-final squared minus mgy.

87
00:05:59,420 --> 00:06:03,370
Let's write that as y-initial.

88
00:06:03,370 --> 00:06:07,340
And we're starting it
at y-initial equals h.

89
00:06:07,340 --> 00:06:10,280
And so, even though
this is complicated,

90
00:06:10,280 --> 00:06:14,180
I can now solve for how much
this spring has been compressed

91
00:06:14,180 --> 00:06:15,672
with a little bit of algebra.

92
00:06:15,672 --> 00:06:17,630
And so I'm just going to
bring a bunch of terms

93
00:06:17,630 --> 00:06:19,280
over to the other side.

94
00:06:19,280 --> 00:06:26,600
And take the square root of 2,
divided by k of mgy-initial.

95
00:06:30,230 --> 00:06:33,126
Minus mu-0 mgd.

96
00:06:35,659 --> 00:06:42,300
Minus mu-1 mgd-squared over 2.

97
00:06:42,300 --> 00:06:46,370
And that's how much the
spring is compressed.

98
00:06:46,370 --> 00:06:49,340
Notice what I did not
do was divide this up

99
00:06:49,340 --> 00:06:51,110
into a bunch of
different motions.

100
00:06:51,110 --> 00:06:53,090
I picked an initial state.

101
00:06:53,090 --> 00:06:54,830
I picked a final state.

102
00:06:54,830 --> 00:06:58,310
I drew my energy diagrams
with my zero points.

103
00:06:58,310 --> 00:07:01,750
I described the key parameters
of initial and final states,

104
00:07:01,750 --> 00:07:04,630
y-i and u-final.

105
00:07:04,630 --> 00:07:08,060
I defined the initial
mechanical energy,

106
00:07:08,060 --> 00:07:10,280
the final mechanical energy.

107
00:07:10,280 --> 00:07:12,670
And then applied
the work energy,

108
00:07:12,670 --> 00:07:14,990
work mechanical
energy principle.

109
00:07:14,990 --> 00:07:17,030
I had to integrate the
friction force because it

110
00:07:17,030 --> 00:07:21,800
was non-trivial and solve
for how much the spring has

111
00:07:21,800 --> 00:07:24,130
been compressed.