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Force vs. Time on a Dynamics Cart - YouTube
Channel: Flipping Physics
[0]
- Good morning.
[1]
Today we're going to
analyze the force of tension
[2]
as a function of time in a string attached
[4]
to a dynamics cart.
[7]
Hey guys.
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- [Billy] Hey Bo.
[8]
- [Bobby] Hi Bo.
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♫ Flipping Physics ♫
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- This is an extension of
a lesson we did previously
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where we solved for the acceleration of
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and force of friction
acting on a dynamics cart
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and we are going to use
some of the information
[23]
from that lesson.
[24]
We know the cart has a
mass of 0.613 kilograms.
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The mass hanging has a
mass of 0.0550 kilograms.
[33]
The acceleration of the
cart after I release it
[35]
is 0.786389 meters per second squared
[39]
and we have a graph of
the force of a tension
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as a function of time.
[47]
There are three main parts of the motion
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we are going to analyze.
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Part one is highlighted in blue
[53]
and is before I release the cart
and the system is at rest.
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During part one, the tension
force is nearly constant.
[61]
Part two is highlighted in red
[62]
and is after I release the cart
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while the system is accelerating.
[66]
For part two, the tension
force decreases a bit
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and then remains rather constant
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while the system is accelerating.
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Part three is highlighted in yellow
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and is while I am catching
the cart to slow it down.
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The tension during part
three increases quite a bit
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and is not constant.
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Now, before we can
analyze this situations,
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we need to draw free dody diagrams.
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Notice there are actually
two free body diagrams
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because there are two
objects that are moving:
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the cart and the mass hanging.
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So we need to draw free body diagrams
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for both the cart and the mass hanging.
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Bo, could you give me all the forces
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and their directions in
both free body diagrams
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and why don't we do it during part two
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while the system is accelerating?
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- [Bo] Sure, the forces on the
cart are;, the force of tension
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to the right in the
direction of the string,
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the force of gravity on
the cart straight down,
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the force normal is
perpendicular to the track
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and straight up.
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The force of friction is
opposite the direction
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the cart is moving, so the force
of friction is to the left.
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The forces on the mass hanging are;
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the force of tension in the direction
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of the direction of the
string, which is up.
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And the force of gravity
on the mass hanging,
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which is down.
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- I have several things to point out.
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First, we have two
different forces of gravity,
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so we've identified them by subscripts:
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c for cart, and h for mass hanging,
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so we have the force of
gravity acting on the cart,
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and the force of gravity
acting on the mass hanging.
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Second, the free body
diagrams don't change much
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from parts one to two to three.
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The only difference is that
during parts one and three
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there's a force applied by
me on the cart to the left.
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- [Mr. P] Lastly, the force of tension
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in these two free body
diagrams has the same magnitude
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because it is the same string
and the pulley has both
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mass and friction that are
so small they are negligible.
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This won't always be true,
but it is true today.
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Before we can start
analyzing the situation
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using Newton's Second Law, we
need to identify a direction.
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Notice how the cart and
mass hanging move together.
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This is because they are
attached by the string.
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Therefore, the cart and
mass hanging have the same
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acceleration and therefore we
call the cart and mass hanging
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"The System".
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Normally the directions
would be; the cart moves
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to the right, which is positive
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and the mass hanging moves
down, which is negative.
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However, because they
have the same acceleration
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they should have the same sign as well.
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So let's identify the
direction the system moves
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as positive.
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You can see I added a curved
arrow with a positive sign
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to show this.
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We will call this the String Direction.
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Now, we are going to
use Newton's Second Law.
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Remind me, Billy, what
is Newton's Second Law?
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- [Billy] Net force equals
mass times acceleration
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where both force and
acceleration are vectors.
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- Let's sum the forces on the mass hanging
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in the positive direction during part one.
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Bobby, could you please sum
the forces on the mass hanging
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in the direction of the
string during part one,
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which is before I release
the cart and then solve
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for the tension force?
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- [Bobby] The net force
on the mass hanging
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in the string direction
during part one is equal to...
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Well the force of tension
is positive because it's up
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and the force of gravity is
negative because it's down,
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so we get the...
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- [Bo] Hold up, Bobby.
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Remember the direction of
the string we just defined?
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- [Bobby] Oh yeah. Thanks.
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- [Bo] You're welcome.
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- [Bobby] So then the net
force equals positive force
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of gravity on the mass hanging
minus the tension force
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which equals the mass
hanging times acceleration.
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To solve for the tension
force, add the tension force
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to both sides and then
subtract the mass hanging
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time acceleration and we get
tension force equals the force
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of gravity on the mass
hanging minus the mass hanging
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times acceleration and we can
substitute in the equation
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for the force of gravity which
is the mass hanging times
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the acceleration due to gravity.
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We can take out the mass hanging
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because it's the common factor,
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and we get the tension force
equals the mass hanging
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times the quantity
acceleration due to gravity
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minus the acceleration of the system,
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which is 0.055 times
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9.81 minus
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the acceleration.
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- [Billy] The acceleration
is zero during part one
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because the system is at rest.
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- [Bobby] Oh yeah. Thanks.
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- [Billy] You are welcome.
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- [Bobby] So it's just 9.8 minus zero,
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which is 0.53955
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or with three sig figs, 0.540 Newtons.
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- [Mr. P] The tension
force of 0.540 Newtons
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is our theoretical prediction.
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Let's see if it matches our
experimental observation.
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The average measured tension
force during part one
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is 0.540 Newtons.
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So for part one, the
theoretical prediction
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matches the experimental
observation, which is good.
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Now let's move on to part two
after I let go of the cart.
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Billy, could you please
solve for the tension force
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in the string during part
two by summing the forces
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on the mass hanging in the
direction of the string
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during part two?
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- [Billy] Well, the free
body diagram is the same,
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so then all the mass should
be the same and we should get
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the same value for the
tension force, right?
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- [Mr. P.] No, okay.
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So the free body diagram
on the mass hanging
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is the same during part two
as it is during part one;
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however, it will not yield
the same force of tension
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when we solve part two.
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So please work your way through
the solution for part two
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to figure out why.
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- [Billy] Okay, well if the
free body diagram is the same
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then everything should be the same.
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Oh wait. Everything is the
same except the acceleration.
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The acceleration is no longer zero.
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It is 0.786389 meters per second squared.
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So then tension equals 0.55
times the quantity 9.81
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minus the 0.786389
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which works out to be 0.496299
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with three sig figs, 0.496 Newtons.
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- [Mr. P] Again, we can compare
the theoretically predicted
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value for the force of
tension of 0.496 Newtons
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to the experimentally
observed value and you can see
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that they do not quite match.
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The experimentally observed
value is 0.490 Newtons,
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which differs by only six thousandths of a Newton,
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which I think is pretty close.
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Let's analyze the third part
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which is while I am stopping the cart.
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The third part is different
from the first two
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because the tension force is not constant.
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So let's solve for something different.
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We know the maximum force
of tension is 0.907 Newtons,
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so let's use that value to determine
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the maximum acceleration of
the mass hanging and cart
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while I am slowing the system down.
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- Bo, could you please do that?
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- [Bo] Well, we need to sum the forces so,
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net force equals the...
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- [Mr. P] Bo, please remember to identify
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all the information about what
you're summing the forces on.
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- [Bo] Okay, let's sum the
forces on the hanging mass
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in the direction of the
string and during part three.
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The equation starts out the same;
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however, divide by the hanging
mass and the acceleration
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equals the force of
gravity of the hanging mass
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minus tension divided by the hanging mass.
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Substitute in the equation for
force of gravity and numbers,
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you get 0.055 times 9.81
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minus 0.907 all divided by 0.055,
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which is negative 6.680909
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or negative 6.68 meters
per second squared.
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- [Mr. P] It is important to
point out that the acceleration
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that we just found is
the maximum acceleration
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which is an instantaneous value.
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Unlike the average values
we figured out in part one
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and part two for the force of tension.
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So instantaneous acceleration,
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the acceleration at a specific point,
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the average force of tension,
the force, average force,
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over a time period.
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In addition to that, the
acceleration we found out
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was negative, which simply
means that the acceleration
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was opposite of the direction
the system was moving
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so it is slowing the system down,
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so technically we
figured out the magnitude
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of the maximum acceleration,
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which was positive 6.68
meters per second squared.
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Thank you very much for
learning with me today.
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I enjoyed learning with you.
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