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course Phy 121
`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)?
The velocity would be 4 and the clock time would be 12.
@& You need to include the units with the quantities (cm/s for velocity and sec for time, as specified at the beginning)*@
@& For velocity vs. clock time the first coordinate is clock time, the second coordinate is velocity. Keep that in mind for the future.
For the purposes of this problem we'll take it as you have interpreted it.*@
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• What are the velocity and clock time corresponding to the point (9, 32)?
The velocity would be 9 and the clock time would be 32.
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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).
The ball will increase its velocity between the points 4 and 9. The velocity will be slower at 4 and faster at 9.
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• What is the change in velocity between these two events?
The change in velocity would be 5 because 4 the initial velocity subtracted from 9 the final velocity is 5m/s.
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• What is the change in clock time between these two events?
The change in clock time between the two events is 20, because 12 the initial clock time subtracted from the final clock time that is 32 equals 20s.
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• What is the average velocity for the interval between these two events?
The average velocity is .25 m/s^2. I used the formula the change in A over the change in B substituting 5 in for A and 20 in for B getting the answer .25.
@& If you have two velocities, you average them by adding them and dividing by 2.
You have calculated the average rate of change of velocity with respect to clock time. This is the correct solution to the next question, but it's not the average velocity.*@
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• What is the average rate at which the velocity changes, with respect to clock time, between these two events?
The velocity will be changing every second .25 m/s^2 is just the average velocity it will start off slower than this and keep changing as long as it is rolling and actually go faster than the .25 m/s^2.
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• What is the displacement of the object between these two events?
I do not know how to do this problem and any assistance would be greatly appreciated.
@& Since ave velocity = change in position / (change in clock time), it follows that
change in position = ave velocity * change in clock time.
You didn't get the average velocity (see previous note), which for your interpretation would be (4 m/s + 9 m/s) / 2 = 6.5 m/s).
Had you gotten that with the units, you would likely have seen that it would be multiplied by your 20 sec time interval to get the change in position, or displacement.*@
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`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
The net force of the object is 10kg m/s^2. I found this by multiplying 2 and 5 to get 10.
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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?
The acceleration of the mass is 50m/s^2. I found this by dividing 5000 by 100 to get 50.
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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?
The work that is done on the automobile is 40,000 Newton meters. I found this by multiplying 400 by 100 getting 40,000.
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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?
The force required for 5 cm is 2.5 Newtons, 10 cm would be 5 Newtons, and for 15 cm it would be 7.5 Newtons.
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• What is the average force required between pullbacks of 5 cm and 15 cm?
The average force that is required for pullbacks between 5 cm and 15 cm is 5 Newtons.
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• How much work is done between these two positions?
The work that is done between these two positions is between 12.5 Newton cm and 112.5 Newton cm. I found this by taking the force required to pullback and multiplying it by the distance it was pulled back.
@& You would multiply the average force on this interval by the displacement.
You've got the right ave force, 5 Newtons. The displacement is from 5 cm to 15 cm. So how much work is involved?*@
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`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)
The average rotation is 3 rotations per second. I found this using the average ROC formula, change in A with respect to change in B. A being the rotations, and B being the change in clock time 4 seconds, giving me the answer 3.
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• Is it speeding up or slowing down?
The system starts out with a velocity of 0, then it starts and speeds up with the average velocity being 3 rotations per second, then it starts slowing down all the way back to a stop and a velocity of 0.
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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).
I found the rate it is speeding up to be 3 rotations per second. I used the formula change in A( rotations) over the change in B(change in clock time). Subtracting the final from the initial (12-0) I got 12 rotations on top and using the clock time on the bottom I got 4 seconds. Dividing this out I found the answer to be 3 rotations per second.
@& This is the average rate at which the system is rotating, not the average rate at which it is speeding up or slowing down.
How fast is it rotating on the average?
How fast at the beginning?
How fast at the end?
So how quickly is that rate of rotation changing?
Not an easy series of questions, but give it a shot.*@
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`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?
@& Remember that KE = 1/2 m v^2. So you can figure out the initial KE.*@
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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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`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).
The kinetic energy changed by -5,000 kg m^2/ s^2 during the interval.
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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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@& Remember, the change in KE is equal to the work done by the net force.*@
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• 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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`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.
Upward as the positive direction.
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• Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
The gravitational force on this object is -20,000 gram cm / s^2. I found this by subtracting the tension force from the gravitational force.
@& That is the net force. You're a step ahead of the questions.
The gravitational forcce is -200 000 g cm/s^2, as given and with the correct sign inserted. The tension force is +180 000 g cm/s^2.*@
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• Using + for your positive direction and - for your negative direction, what is the tension force on this object?
The tension force is actually -20,000 gram cm / s^2 also. It has the (-) sign in front of it but because the full gravitational force is greater than tension force it cancels the tension force out, but the tension force does take away the majority of the gravitational force.
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• Using + for your positive direction and - for your negative direction, what is the net force on this object?
The net force of the object is -20,000 gram cm / s^2. Because the net force is just the total of all the forces acting on an object.
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• Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
This object does not have any acceleration because it is not moving. I have looked through all my notes and every acceleration formula requires that it have a velocity and a clock time that it moved.
@& Net force = mass * acceleration.
You have correctly figured out the net force and you have been given the mass.
So you can use that formula to find the acceleraiton.*@
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• 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?
The displacement is upward.
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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.
My result is - 6,000,000 gram cm^2 / s^2.
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• When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
My result is positive 5,400,000 gram cm^2 / s^2.
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• When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
My result is -600,000 gram cm^2 / s^2.
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• Does gravity do positive or negative work on this object?
Gravity does negative work on this object.
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• Does the tension force do positive or negative work on this object?
The tension force does positive work on this object.
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• Does the net force do positive or negative work on this object?
The net force does negative work on this object.
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• Does the object speed up or slow down?
The object slows down.
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• How would your answers have changed if you had chosen the opposite direction as positive?
All of my positive answers would have negative and all of my negative answers would have positive.
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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?
The force the pendulum has at .1m is 1.5 kg / s^2 * m.
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• How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
At .05m its force is .75 kg / s^2 * m, at .1m it is 1.5 kg / s^2 * m, at .15m it is 2.25 kg / s^2 * m, and at .2m it is 3 kg / s^2 * m.
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• What do you think is the average force between | x | = .05 meter and x = .2 meter?
I think the average force between .05m and .2m is 1.875 kg / s^2 * m.
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• How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
@& You know the average force.
work = ave force * displacement
How much displacement is there between the two given points?
So how much work?*@
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• 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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• 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?
The direction of the force would be in the same direction as the motion. Because the pendulum is moving in the same direction as it moves from .5m to .2m.
@& From .05 m to .2 m the pendulum moves away from equilibrium.
The force is directed back toward equilibrium.*@
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• 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?
The direction of the force would be in the opposite direction of the pendulums motion. Because it looks like .20m is its maximum height on that swing and it is moving in the opposite direction.
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• If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
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• 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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• When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
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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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• What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
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`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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• 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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• What is the run between these points?
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• What is the average slope between these points?
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• What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
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• What is the width of the trapezoid?
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• What therefore is the area of the trapezoid?
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• What are the graph points corresponding to x = .05 meter and to x = .20 meter?
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• What is the area of the 'graph trapezoid' defined by these points?
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• What is the meaning of the altitude of this trapezoid?
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• What is the meaning of the width of this trapezoid?
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• What therefore is the meaning of the area of this trapezoid?
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`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?
The average rate of rotation when the weights were at the end of the beam was .4 rotations per second.
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• What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
The initial rate of rotation would have been 2.5 rotations per second.
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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?
At that instant they were moving at rate of 140 cm / second.
@& At 2.5 rotations / second , since the object would be moving a distance equal to the circumference of the circle every second, it would be moving at 2.5 times the circumference every second.*@
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• Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
The kinetic energy of each mass is 5,625 gram cm/second. Using the formula I plugged my numbers into it getting (1/2)(60g)(187.5 cm / second)^2.
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@& Good work on most questions.
Check my notes for some corrections and clarifications.*@