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Assignment 14 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.
• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) how much work is required to stretch it?
answer/question/discussion: If the tension in the rubber band is 100 % conservative, we know that the work done on the system is cancele out by the work done by the system. Therefore, the work required to stretch the rubber band will have a net force of 0 N, but if stretched to maximum of 3 N, for example, there is a -3N force canceling it out.
Work is the product of the force in the direction of the displacement and the displacement.
To stretch the rubber band requires a displacement of the one or both ends, with stretching forces exerted in the directions of any displacements. So to stretch the rubber band requires positive work.
The total displacement is 10 cm - 8 cm = 2 cm.
The force changes linearly with displacement, from 0 N to 3 N, so the average force is (0 N + 3 N) / 2 = 1.5 N.
The work is therefore force * displacement = 1.5 N * 2 cm = 3.0 N * cm, or .03 N * m = .03 kg m^2 / s^2, or .03 Joules.
1.5 N * 2 cm is also 150,000 g cm/s^2 * 2 cm = 300,000 g cm^2 / s^2, or 150,000 dynes * 2 cm = 3 * 10^5 ergs.
• During the stretching process is the tension force in the direction of motion or opposite to the direction of motion?
answer/question/discussion: During the stretching process, the tension force in the direction of motion is opposite and in the direction of motion. We know this because the force of the system is a net of 0 N.
• Does the tension force therefore do positive or negative work?
answer/question/discussion: The force of the tension therefore does both positive and negative work, since it is abiding by a conservative force system and has 0 N net work.
The tension force is directed opposite the displacement of either of the ends. So as the rubber band is stretched the tension force does negative work.
If we stretch the rubber band, our forces are in the direction of the displacement of the ends, so we do positive work.
The work done by the stretching force and the work done by the tension force are equal and opposite.
The rubber band is released and as it contracts back to its 8 cm length it exerts its tension force on a domino of mass .02 kg, which is initially at rest.
• Again assuming that the tension force is conservative, how much work does the tension force do on the domino?
answer/question/discussion: If we assume that the acceleration is 980 cm/s^2, and we know that 10cm – 8cm = 2 cm as ‘ds, we cancalculate Fnet= m * a = .02 kg * 980 cm = 19.6 N.
Gravity isn't directly involved in the acceleration of the domino. If you stretched out the same rubber band on the space station, on the moon, on a spaceship traveling far from any star or planet, the acceleration of the domino would be pretty much the same.
980 cm/s^2 is the acceleration of gravity near the surface of the Earth, and is not related to the action of this rubber band.
The work done to stretch the rubber band is about .03 Joules or 3 * 10^5 ergs. The work done by tension as the rubber band is stretched is equal and opposite to this.
As the rubber band returns to its original length, the tension force does positive work. In the ideal case the tension forces will be equal and opposite to those experienced during stretching, so the tension does work equal and opposite to the -.03 Joules of the stretching process. The tension will therefore do +.03 Joules of work on the domino.
In reality the tension forces exerted during contraction are not identical to those exerted during the stretching, and the work done on the domino would be a bit less than .03 Joules. However in the ideal case the work will be .03 Joules.
• Assuming this is the only force acting on the domino, what will then be its kinetic energy when the rubber band reaches its 8 cm length?
answer/question/discussion: If this is the only force acting on the dominoe, the kinetic energy will then be at 0 joules when its 8 cm length is reached. This is because it is now back at its restful stage and is no longer moving.
Having been subjected to a nonzero net force, which has done .03 Joules of work on it, the domino will be moving.
It started from rest and .03 Joules of work were done on it, so its kinetic energy will be .03 Joules.
Whether the solution is + or - depends on your original choice of the positive direction.
• How fast will the domino be moving?
answer/question/discussion: The domino will be moving at 62.72 m/s at its final velocity. This was found by a series of equations. First, we can use the equation y= ˝ * a * ‘dt^2 to find ‘dt (because we know initial velocity is = to 0 cm/s). And we get a value of .064 sec for ‘dt. Once we know this, we can use the equation vf= a*’dt+v0 to solve for vf. We get vf= 980 cm/s^2 * 0.064 sec + 0 cm/s = 62.72 cm/sec.
Gravity has no significant effect on this system. This interaction is between the rubber band and the domino.
KE = 1/2 m v^2. Rearranging this equation to solve for v we have
v = +-sqrt( 2 KE / m). Substituting the mass of the domino and the KE we have
v = +-sqrt( 2 * .03 J / (.02 kg) )
= +- sqrt ( 3 J / kg)
= +- sqrt( 3 (kg m^2/s^2) / kg)
= +- sqrt( 3 m^2/s^2) = +- 1.7 m/s.
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It took me about 30 minutes to do this work.
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You're doing some good thinking, but the acceleration of gravity mislead you, and you need to be sure you apply the definition of work. I've inserted several notes. Be sure you understand the solutions.
&#Let me know if you have questions. &#