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course PHY 231
Sorry so late, got off work at 12 and had to copy everything into this document. I still have some questions though. The ones that I have a bunch of question marks on are the ones that stumped me.
Applying basic definitions of motion, force, work and energy to some of our fundamental systems
A lot of this document can be regarded as sort of a reading comprehension exercise.
If you interpret the words right you can get the right answers, even if you're not completely sure what some of the answers mean.
The main task is to figure out what to do with the given information, in terms of the relationships you have been given, and do the appropriate calculations (being sure to include the units; however you aren't asked to interpret the units and you may simply express the units as they are given).
Another important task is to get used to the ideas of motion, force, work-energy, etc. in the context of some of the systems we have observed.
Information from preceding assignments:
Rate of change:
The average rate of change of A with respect to B is (change in A) / (change in B).
Average velocity is average rate of change of change of position with respect to clock time.
Average acceleration is average rate of change of position with respect to clock time.
Graph Trapezoids
The slope of a 'graph trapezoid' is its rise / run.
The 'graph altitudes' of a 'graph trapezoid' are the quantities represented by its vertical sides.
The area of a 'graph trapezoid' is its average 'graph altitude' multiplied by its width.
Force, work-energy, momentum
When an object of mass m is moving with velocity v, it has the following properties
its kinetic energy is KE = 1/2 m v^2
its momentum is p = m v
Forces acting on objects can change their velocity, momentum and kinetic energy.
When an object of mass m changes its velocity, with respect to clock time, at rate a, then the net force acting on it (i.e., the sum of all the forces acting on it) is F_net = m * a.
If a force F acts through a displacement `ds along the line of the force, then the force does work `dW = F * `ds.
If F in the above happens to be the net force acting on an object, then the KE of that object changes by an amount equal to `dW.
If a net force F_net acts on an object for time interval `dt, then the momentum of that object changes by `dp = F_net * `dt.
As on the preceding assignment, some questions can be answered fairly directly, while others are more challenging. Don't let yourself get bogged down on any one question before moving on to another. Come back on another day to questions you can't answer on your first try.
`q001. On a graph of velocity v (in cm/sec) vs. clock time t (in sec):
What are the velocity and clock time corresponding to the point (4, 12)?
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Velocity= 12 cm/sec
Clock Time= 4 sec
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What are the velocity and clock time corresponding to the point (9, 32)?
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Velocity=32 cm/sec
Clock time= 9 sec
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If these points correspond to the velocity of a ball rolling down an incline, describe as fully as you can what you think happens between the first event (corresponding to the first point of the graph) and the second event (corresponding to the second point of the graph).
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If these two points were describing a ball rolling down an incline, then it would show me that as time elasped, the velocity of the ball was speeding up. Also, by the time the ball reaches the end of the ramp, it's velocity is 32 cm/sec.
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What is the change in velocity between these two events?
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The change in velocity is from 32 cm/sec to 12 cm/sec which is 32-12 which equals 20 cm/sec
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What is the change in clock time between these two events?
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The change in clock time is from 9 seconds to 4 seconds which is 9-4= 5 seconds
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@& Minor point but important:
The velocity goes from 12 cm/s to 32 cm/s and the clock time from 4 sec to 9 sec, the reverse of your statement.
You did calculate the changes in velocity correctly.*@
What is the average velocity for the interval between these two events?
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The V ave is going to be the Vf +V0/ 2= V ave
32cm + 12cm =44cm /2
22 cm
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@& Those velocities are in cm / sec. So your sum and your average will also be in cm / sec.*@
What is the average rate at which the velocity changes, with respect to clock time, between these two events?
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The average velocity is the change in a with respect b which is 20 cm / 5 sec which is 4 cm/sec
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@& Almost. The change in velocity is 20 cm/s, not 20 cm. The change in clock time is 5 s, so the average rate of change is (20 cm/s) / (5 s) = 4 cm/s^2.*@
What is the displacement of the object between these two events?
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The displacement of the objects is the change in position.
The starting position is 12 cm and the ending position is 32 cm.
Therefore, the displacement is 32cm-12cm= 20cm
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@& vAve = (change in position) / (change in clock time), so
(change in position) = vAve * (change in clock time
You found a correct average velocity, 22 cm/s (though you didn't quite have the units right) and you know the time interval, so what do you conclude is the change in position?*@
`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
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Net force is the total amount of force acting on an object. Net force = m*acceleration
5 kg * 2 m/s^2= 10 kg* m/s^2
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`q003. A net force of 5000 kg m/s^2 acts on a 100 kg mass. What is the acceleration of the mass?
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To find the acceleration of when given the net force you will have to apply the net force formula which is m*acceleration
5000 kg m/s^2= 100 kg * a
5000/100= 50 m/ s^2
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`q004. A force of 400 Newtons is exerted on an automobile as it is pushed through a distance of 100 meters. How much work is done on the automobile?
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400 N * 100 m= 40,000 joules
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`q005. A certain pendulum requires .5 Newtons of force for every centimeter it is pulled back (recall pulling back the pendulum hanging from the tree limb, using the rubber band).
How much force would be required to pull the pendulum back 5 cm, 10 cm and 15 cm?
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5 cm= .5 N * 5 cm= 2.5 N*cm
10cm= .5 N * 10 cm= 5 N*cm
15cm= .5 N * 15 cm= 7.5 N*cm
@& Good, but think about the units:
At .5 N for every centimeter, a displacement of 5 cm will give you 2.5 N, not 2.5 N * cm.
.5 N for every cm could be written .5 N / cm, which when multiplied by 5 cm gives you 2.5 N.*@
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What is the average force required between pullbacks of 5 cm and 15 cm?
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7.5 N*cm + 2.5 N*cm /2= 5 N*cm
@& Good, but it's just 5 N.*@
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How much work is done between these two positions?
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Im not sure on this answer but here goes:
If the average force being done = 5N*cm and the the ave in between 15cm and 5cm is 10 cm, then the work being done is 5N cm * 10 cm= 50 N*cm^2
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@& Good.
If you use 5 N instead of 5 N * cm, you get 50 N * cm.*@
`q006. A system rotates through 12 rotations in 4 seconds, first coming to rest at the end of this interval.
How quickly is it rotating, on the average? (The answer is as obvious as it should seem, but also be sure to interpret this as a rate of change with respect to clock time, and carefully apply the definition of average rate)
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If the average rate with respect to clock time is change in a divided by the change in b then it would be 12 rotations/ 4 seconds which would be 3 rotations every second
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Is it speeding up or slowing down?
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If there is no force applied other than itself and gravity, it will be slowing down.
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At what rate is it doing so? (Again attempt to interpret as an average rate of change of an appropriate quantity with respect to clock time).
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Every 3 rotations
if it only did 12 rotations in 4 seconds, the ave rate of change will mean that it is slowing down because the force is changing every 3 roations.
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@& 3 rot / sec is the average rate at which it's rotating.
The question is the tough one: By how much does its rate of rotation change?
You know that its average rate of rotation is 3 rot / sec and its final rate is 0, so what are its initial and final rates of rotation?
How much slowing down does it therefore do on this interval, and how long does it take to do so?
At what average rate is it therefore slowing?*@
`q007. At the initial point of an interval a 7 kg mass is moving at 5 meters / second. By the end of the interval it has gained an additional 200 kg m^2 / s^2 of kinetic energy.
How much kinetic energy does it therefore have at the end of the interval?
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KE=1/2mv^2
3.5kg (5m/sec)^2= 3.5 kg (25m/sec^2)= 87.5 kg m^2/sec^2
200 kg m^2/sec^2 + 87.5 kg m^2/sec^2 = 287.5 kg m^2/sec^2
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Slightly more challenging question: How fast is the mass therefore moving at the end of the interval?
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Ok, here goes:
If it is moving 287.5 kg m^2/sec^2 then you would have to apply the rate of change definition I think...
so it would be 287.5/87.5 ~ 3.3 m/sec
Then you would multiply maybe????
5 m/sec * 3.3 m/sec= 16.5 m/sec
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@& Good try, but what you need to do is set 1/2 m v^2 = KE and solve for v. You know KE and m, so you can get a solution for v.
We'll go over the algebra of this solution soon.*@
`q008. An object begins an interval with a kinetic energy of 20 000 kg m^2 / s^2, and ends the interval with a kinetic energy of 15 000 kg m^2 / s^2.
By how much did its kinetic energy change on this interval? (The answer is as obvious as it might seem, but be careful about whether the answer is positive or negative).
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20,000 kg m^2/s^2 - 15,000 kg m^2/s^2= 5,000, however since it is slowing down it is losing KE which means it is -5 000 kg m^2/s^2
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More challenging: During this interval, how much work was done on the object by the net force?
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Work is defined as the total amount of work put on an object and the net force is m*a ????????????
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@& Note that the change in KE is equal to the work done by the net force.*@
Also more challenging: If the average force on the object during this interval had magnitude 200 Newtons, then what was its displacement during this interval?
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??????????????????????
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@& The work done is equal to force * displacement.
Once you figure out the work done (see previous notes) you can use this with the given displacement to find the force.*@
`q009. A mass of 200 grams hangs from one side of a pulley, and another mass from the other side. The gravitational force pulling down on this mass is about 200 000 gram cm / s^2, and the tension in the string pulling it upward is about 180 000 gram cm / s^2.
Pick either upward or downward as the positive direction.
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downward
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Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
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+
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Using + for your positive direction and - for your negative direction, what is the tension force on this object?
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-
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Using + for your positive direction and - for your negative direction, what is the net force on this object?
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+
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@& The question asked for the net force, not just the sign of the net force.
You combine the gravitaional and tension forces to get a net force of +20 000 g cm / s^2, based on your choice of downward as the positive direction.*@
Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
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+
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@& You can find the magnitude of the acceleration as well.
F_net = mass * acceleration
You are given the mass and you know F_net. So what is the acceleration?*@
If the object's displacement during a certain interval is +30 cm, then according to your choice of positive direction, is the displacement upward or downward?
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downward
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When you multiply the displacement by the gravitational force, what is your result? Be sure to indicate whether the result is positive or negative.
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(displacement) 20,000 g cm/sec^2 * (Grav F) 200,000 g cm/sec^2= 4,000,000,000 g cm/sec^2 ??????????????
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@& The displacement is -30 cm, as you will understand from your previous answers.
This is what you would multiply by your gravitational force.*@
When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
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20,000 * 180,000= 3,600,000,000 g cm/sec^2 ?????????????
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@& Again the displacement is -30 cm,.*@
When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
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200 g * 200,000 g cm/sec^2= 40,000,000 g cm/sec^2 gives you net force
@& This would be mass * grav force, not displacement * net force.
This idea also applies below.*@
40,000,000 g cm/sec^2 * 20,000 g cm/sec^2= 800,000,000,000 g cm/sec^2
Positive
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Does gravity do positive or negative work on this object?
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+
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Does the tension force do positive or negative work on this object?
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Negative
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Does the net force do positive or negative work on this object?
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Positive
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Does the object speed up or slow down?
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Speed up
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How would your answers have changed if you had chosen the opposite direction as positive?
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Everything would be completely opposite
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`q010. A pendulum of length 2 meters and mass 3 kg, pulled back a distance | x | from its equilibrium position, experiences a restoring force of magnitude k | x |, where k = 15 kg / s^2 * | x |. [Note that for convenience in calculation we are making some approximations here; the actual value of k for this pendulum would actually be closer to 14.7 kg / s^2, and this is so only for values of | x | which are a good bit smaller than the length. These are details we'll worry about later.]
How much force does the pendulum experience when x = .1 meter?
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15 kg / sec^2 * .1 m = 1.5 m kg/sec^2
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How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
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15 kg /sec^2 * .05m = .75 m kg/sec^2
15 kg / sec^2 * .1m= 1.5 m kg/sec^2
15 kg /sec^2 * .15 m = 2.25 m kg/sec^2
15 kg/ sec^2 * .2 m = 3 m kg/sec^2
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What do you think is the average force between | x | = .05 meter and x = .2 meter?
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Average Rate of Change definition is used with this problem
3 m kg/sec^2 + .75 m kg/sec^2 = 3.75 m kg/sec^2 /2 = 1.875 kg/sec^2
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How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
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1.9 kg/sec^2 * 3 kg= 5.7 kg^2/sec^2
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@& You've got the average force, in a previous answer.
What is the displacement between these positions?
How do you use average force and displacement to get work?*@
How fast would the pendulum have to be going in order for its kinetic energy to equal the result you just obtained for the work?
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3,000 kg * 1.875 kg/ sec = 5625 kg ^2 / sec or 5.625 g^2 / sec
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@& Here you would set 1/2 m v^2 equal to the work done, and knowing m you could solve for v.
I don't necessarily expect that everyone will recall how to do the algebra, and we'll be spending some class time on that.*@
If the pendulum moves from position x = .05 meter to x = .2 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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The same
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@& The force acts back toward the equilibrium position at x = 0, the displacement is away from equilibrium, so the two have opposite directions.*@
If the pendulum moves from position x = .20 meter to x = ..05 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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Opposite
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@& Consider my previous note and how it would be adapted to this question.*@
If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
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.05 m / .75 m kg/ sec^2= .07/ 1 kg
.07 kg / sec^2
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@& If the string is cut the string stops exerting a force and the pendulum is subject only to the force of gravity, which accelerates it downward at 9.8 m/s^2.*@
What is the magnitude of the force exerted by gravity on the pendulum's mass? For simplicity of calculation you may use 10 m/s^2 for the acceleration of gravity.
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force of magnitude is k * x
k= 10 m/ s^2 (3 g)
30 g m/ s^2
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@& The mass is 3 kg, so you would want to replace your g with kg.*@
When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
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.1 m/ 2 m = .05 which is 5%
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When x = .1 meter, what is the restoring force as a percent of the pendulum's weight?
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15 kg/ sec^2 * .1m= 1.5 m kg/ sec ^2
3 kg/1.5 m kg/ sec^2
50%
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@& The weight is 3 kg * 10 m/s^2 = 30 kg m/s^2.
You should recalculate this and compare with your answer to the preceding question.*@
What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
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If the restoring force of magnitude is K lxl, then I would think that the magnitude of acceleration would be the opposite.
so, k/x
15 kg / sec^2 / .15 m= 100 m kg/sec^2
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@& Nice try, but consider:
You previously found the force on the pendulum at this point.
You know the mass of the pendulum.
So you can use F = m a to find the acceleration.*@
`q011. The force exerted on a mass has magnitude | F | = 15 Newton / meter * x, for 0 <= x <= .25 meter.
Sketch a graph of | F | vs. x. (You might wish to start by making a table of | F | vs. x, for some appropriate values of x between 0 and .25 meter). Note the convention that a graph of y vs. x has y on the vertical axis and x on the horizontal, so that for this graph | F | will be on the vertical axis and x on the horizontal.
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How am I going to sketch a graph on here? However, I did on a sheet of paper.
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@& You can't sketch the graph here. In general you will describe what you see in the graph. Your answers to the next couple of questions will cover this.*@
Verify that the points (.06 meter, .9 Newton) and (.16 meter, 2.4 Newtons) lie on your graph.
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What is the rise between these points?
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2.4N - .9 N= 1.5 N
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What is the run between these points?
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.16 m - .06 m = .1 m
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What is the average slope between these points?
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Ave slope = 1.5N/ .1 m = 15 N * m
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@& Your dividing N by m so the unit is N / m. Otherwise right.*@
What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
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2.4 N + .9 N = 3.3 N
3.3 N / 2 = 1.65 N
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What is the width of the trapezoid?
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.1 meter
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What therefore is the area of the trapezoid?
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1.65 N * .1 m = .165 N * m
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What are the graph points corresponding to x = .05 meter and to x = .20 meter?
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15 N / m * .05 m= .75 N
15 N/ m * .20 = 3 N
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What is the area of the 'graph trapezoid' defined by these points?
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3 + .75N= 3.75/ 2 = 1.875 N
.2 m - .05 m = 1.5 m
1.875 N * 1.5 m = 2.8 joules
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What is the meaning of the altitude of this trapezoid?
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The altitude describes the height of trapezoid and gives us our slope. It describes the force
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@& Altitude and slope are different things.
But an altitude is a force, as you say.*@
What is the meaning of the width of this trapezoid?
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The width describes the amount of distance traveled.
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What therefore is the meaning of the area of this trapezoid?
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The area is the entire amount of force and distance travled
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@& Good.
And when you multiply force by distance you get work.*@
`q012. For the rotation data you took in class:
What was the average rate of rotation in the trial where the added masses were at the end of the rotating beam?
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720 degree , 6 seconds
120 deg/sec
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What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
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initial is going to be 240 deg/sec because 120 deg/sec is the Vave
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Assuming that those masses were 14 cm from the center of rotation, how fast were they moving, in cm / second, at that initial instant?
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Vo is 240 degrees/ sec and
Vave is 120 deg/sec
vf is 0 deg/sec
Im confused on what we do with the cm because we were dealing with degrees and now we are dealing with cm.
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@& Not many people are getting this, so we'll be talking about it.
The thing you use here is the fact that the circumference of a circle is 2 pi r. So each rotation corresponds to a distance equal to the circumference.*@
Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
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KE= 1/2 m v^2
30 (120 deg/ sec) ^2
432000 g * deg^2/ sec^2
???????
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@& You missed a few things, but not that many. You're doing really well with this. Keep up the good work.*@