cq_1_151

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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You haven't given the maximum and minimum tensions.

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.

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.

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 to solve this problem 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.

This energy could also be expressed as 300,000 ergs.

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.

It's not required here, in fact not even advised, 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.

20 grams is .020 kg, not .002 kg. KE is .03 J, not 100 J.

With these values your solution would work. You would get a velocity of about 1.7 m/s.

Good solution, given your value of v.

However if you know that 3 N is less than a pound, and imagine a rubber band that stretches 2 cm in response to this force, it's obvious that it won't shoot a domino 3 miles high.

It's a good idea to solve in terms of energy, with with v0 = 1.7 m/s your solution would give you `ds = .15 m or 15 cm, approx..

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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&#This looks good. See my notes. Let me know if you have any questions. &#