Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Assignment 15 Seed Question 1
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
• Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
answer/question/discussion: Because the mass of the rubber band is negligible, the force exerted at one end is transmitted undiminished to each adjacent piece of the rubber band along the entire length to the other end. Thus, the forces pulling on the rubber band at its two ends must add up to zero. Thus, we would say that because the system’s force must equal zero newtons, we can say that at the maximum tension, 3 Newtons must have been exerted. Therefore, when 3 Newtons is exerted on the object, there must also be an equal and opposite force exerted of -3N.
The force exerted by tension at one end is indeed equal and opposite to the force it exerts at the other.
The tension itself has no direction, but it exerts equal and opposite forces at every point of the rubber band except at the ends.
At each end the tension force toward the other end is unopposed by a tension in the opposite direction (e.g., at the right end there is a tension force toward the left but, unlike at a point between the two ends, there is no equal and opposite force to the right). So at each end the tension exerts a force toward the other end.
The minimum tension during the stretching process is 0 N, at the beginning when the length is 8 cm. The maximum is 3 N, at then end when the length is 10 cm. There is no reason to prefer the 0 N force to the 3 N force; they are both equally applicable to the stretching process. The reasonable conjecture is that the average force is equal to the average of these equally applicable forces, so the average force is (0 N + 3 N) / 2 = 1.5 N.
• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: The elastic potential energy in the rubber band at the 10 cm length will be ˝ * k * x^2, so we need to figure out k. TO find k, we use F=kx, so 3 N = k * 2cm, so k= 1.5. So we know that elastic PE = ˝ * k * x^2 = ˝ * 1.5 * 2cm^2 = 3 joules.
You don't need a formula for elastic potential energy to find this result, and at this point the typical student is not equipped to use this definition. To understand the elastic PE formula you need to understand how this problem is reasoned out in terms of the definitions of work/energy and potential energy, without the use of the formula. This is one of the main goals of this problem.
All that's required is the knowledge that the change in elastic potential energy is the work done against the conservative force in order to stretch the rubber band.
The work done is equal to the product of the average force and the displacement in the direction of the force.
The average force is 1.5 N, the displacement is 2 cm. The force needed to stretch the rubber band is in the direction of the displacement, so we have
change in elastic potential energy = 1.5 N * 2 cm = 3 N * cm.
Notes:
This is an instance of the principle that change in potential energy is defined as work done against a conservative force.
If we take the potential energy of the unstretched rubber band to be 0, then this potential energy change is equal to the potential energy of the stretched rubber band.
N * cm is not a standard unit of energy. A cm is .01 m, so the unit N * cm is N * (.01 m) = .01 N * m = .01 Joules.
The potential energy of this stretched rubber band, relative to its unstretched state, is therefore 3 N * cm = 3 ( .01 Joules) = .03 Joules.
The same result could alternatively have been calculated directly as 1.5 N * .02 m = .03 N * m = .03 Joules.
This energy could also be expressed as 300,000 ergs.
It's not required here but since you mention the elastic PE formula, here's how it can be used to solve this problem. The formula is PE = 1/2 k x^2. In this case x = 2 cm = .02 m.
k is the force constant, defined as rate of change of force with respect to position. In this case the force constant is
k = change in force / change in position = (3 N - 0 N) / (2 cm) = 1.5 N / cm.
Thus 1/2 k x^2 = 1/2 (1.5 N/cm) * (2 cm)^2 = 1/2 * 1.5 N/cm * 4 cm^2 = 3 N cm, as before.
As before it's a good idea to express this in standard units as .03 J or 300,000 ergs.
• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: Kinetic energy is now equal to 3 joules. We know that KE = ˝ m v^2 so we know that 3 joules= ˝ * 20 g * v^2. We can now solve for v^2. We can get the answer by dividing 3 joules by 10 g and then taking the square root of this number to get .578 cm/s.
Good, but your units are wrong and the energy wasn't previously calculated quite correctly.
We have KE = 1/2 m v^2. Solving for v we get
v = sqrt( 2 KE / m) = sqrt( 2 * .03 Joules / (.020 kg) ) = sqrt( 3 J / kg) = sqrt( (3 N m^2 / kg^2 ) / kg ) = 1.7 m/s^2, approx..
• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: To find the distance, we need to use the equation W= ‘dKE. Since we know this, we can say that KE = F * ‘ds, so 3 joules = 3 Newtons * ‘ds and the distance it travels up is going to be equal to 1 m.
The domino was given a kinetic energy of .03 Joules. It will rise until its KE has decreased to 0. At this point is gravitational PE will have increased to .03 Joules.
The force exerted on the domino by gravity is .020 kg * 9.8 m/s^2 = .196 Newtons. So the work done against gravity is .196 Newtons * `dy, where `dy is the displacement in the vertical direction. We could set .196 Newtons * `dy equal to .03 Joules and solve for `dy, obtaining `dy = .15 meter or 15 cm.
We should also be able to solve this situation symbolically:
Symbolically the work done against gravity is equal to its weight m g multiplied by the change `dy in its vertical position, so
`dPE_grav = m g * `dy.
In this situation we know `dPE_grav must be .03 Joules, m = .020 kg, g = 9.8 m/s^2 and `dy is the vertical displacement. Solving the equation for `dy we divide both sides by m g to get
`dy = `dPE_grav / (m g) = .03 J / (.020 kg * 9.8 m/s^2) = .15 J / (kg m/s^2) = .15 (kg m^2/s^2) / (kg m/s^2) = .15 meter.
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I took about 30 minutes on this.
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You did a lot of good thinking on this problem. You didn't always use units appropriately and you did go astray on a couple of the concepts.
See my notes and let me know if you have questions.