Phy 201
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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:
We know from the last seed question the min 1newton and max is 3newtons
The minimum tension is 0, which occurs just before the rubber band begins to exert a force.
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:
Since the force is =3newtons and the `ds is 10cm then 3newtons *.10m=.3joules
Since PE equal to the work done to pull it back then PE=.3joules
`ds for the stretch is not 10 cm. The rubber band starts out at length 8 cm with 0 force.
3 N is the maximum force, and there is no reason to base the calculation of work on the maximum force. The work is the product of the average force and the displacement.
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:
If we use the equation v=sqrt(2*KE/m) we can find v. Assuming the PE will decrease and KE
will increase and be equal to PE.v=sqrt(2*.3joules/.020kg),v=sqrt(.6joules/.020kg),v=5.5m/s
Right procedure. With the correct value for the work this woud give the correct result.
If instead the rubber band is used to 'shoot' the domino straight upward, then how high will
it rise?
answer/question/discussion:
If we take m*g we get the Fgrav which is .020kg*9.8m/s^2=.196newtons If we subtract the
Fgrav by the Ftension we get, Fnet=3newtons-.196newtons,Fnet=2.8newtons. I'm lost.
The .196 Newton weight is just what you need.
The domino will rise until the energy it gains from the rubber band has been lost to the work done on it by gravity--i.e., until its initial KE has been converted to gravitational PE. If nothing gets in the way, it will then fall, converting its gravitational PE back into KE.
Here is a detailed analysis and explanation. As always, you are welcome to submit a copy of this document and insert anything you with for my feedback, marking your insertions before and after with &&&&.
nature of tension force
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.
average tension
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.
solution (elastic PE formulas not needed)
You don't need a formula for elastic potential energy to find this result, and at this point the typical student is not yet 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.
Since Newtons and cm are from different systems of units, it's a good idea to make the units consistent. Since 1 m = 100 cm, we have 1 cm = 1/100 m = .01 m, so that 1 N * cm = 1 N * .01 m = .01 N * m. Similarly
change in elastic potential energy = 3 N * cm = 3 N * .01 m = .03 N * m = .03 Joules.
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.
correct velocity
By definition 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..
domino height
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.
It is preferable to first 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.
alternative calculation of rubber band energy using the elastic PE equation
You should be able to solve this problem based on the definition of PE, as shown above. Its not required here, in fact it's not even advised, but since you might well have encountered the elastic PE formula (in fact some students have attempted to solve the problem using this formula), here's how it can be used to solve this problem.
The formula is PE = 1/2 k x^2, where the PE is measured relative to the equilibrium length of the elastic object and x is the stretch or compression relative to equilibrium configuration. In the case of a rubber band compression is not an issue--compression does not result in an elastic response--but for some springs at many other elastic objects at least a degree of compression is possible. This formula applies to any object for which the elastic force is directly proportional to the stretch or compression, so that the elastic force exerted by the object is F_elastic = - k x. As the object is stretched from its equilibrium length to length x the force required to stretch it changes linearly from 0 to k x, so that the average force is (0 + kx) / 2 = 1/2 k x. The displacement during this stretch is x so that the work required to stretch the object is 1/2 k x * x = 1/2 k x^2. This energy is 'stored' as elastic potential energy, until the object is released and allowed to 'snap' or recoil back to its original position.
The energy of the stretched rubber band could thus have been calculated directly as 1.5 N * .02 m = .03 N * m = .03 Joules, using PE_elastic = 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 its a good idea to express this in standard units as .03 J or 300,000 ergs.
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50min
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