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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.
The tension increases from 0 N at the 8 cm
length to 3 N at the 10 cm length.
If tension is a linear function of length,
then the average tension is (0 N + 3 N) / 2 = 1.5 N.
It is unlikely that a rubber band has a
perfectly linear tension function, but without further information this is
the most reasonable approximation to make, and it probably isn't much in
error.
See further
discussion of tension vs. length, including graphs and the idea of tension
vs. stretch, under Additional Discussion near the end of this document.
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 done by the stretching force is therefore
-
work exerted on rubber band = force * displacement = 1.5 N * 2 cm =
3.0 N * cm, or .03 Joules.
The stretching force is in the direction of
motion and does positive work.
However the tension force is the force
exerted by, not on, the rubber band, and is in the direction opposite
motion. The tension force therefore does negative work during the
stretching process.
More about the tension force:
Tension forces act throughout the rubber
band. If we regard the mass (and therefore the weight) of the rubber band as
negligible, tension exerts equal and opposite forces at every point of the
band, except its ends.
At each end, since the rubber band exists only on one side of that end,
there is no equal and opposite tension force so the tension acts in only one
direction, toward the opposite end.
Thus the tension force at the end attached to the domino acts in the
direction from the attached end to the opposite end of the rubber band.
During the stretching process, this is opposite the direction of motion.
After release, the direction of motion will be toward the opposite end of
the band, so the tension force and displacement will be in the same
direction.
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.
If the force is conservative, then the work
done on return is equal and opposite to the work done in the stretching
process. That work was negative, so the work done on return is
positive.
We conclude that the work done by the
tension force is on return is +.03 Joules
Given the stated assumption, the tension
force is the net force.
The work done by the net force acting on an
object or system is equal to the change in KE of that object or system:
Since `dW_net is in this case +.03 Joules,
`dKE = +.03 J and the domino's KE increases by .03 J.
The domino was originally at rest, so its
KE increases from 0 to .03 Joules.
KE = 1/2 m v^2, so v = sqrt(2 * KE / m).
Thus when its KE is .03 Joules, the .02 kg domino is moving at velocity
v = sqrt( 2 * .03 Joules / (.02 kg) ) =
sqrt(3 m^2 / s^2) = 1.7 m/s, approx.
STUDENT ERROR (plus commentary on F = - k x):
F = kx
3 = k10
k = 0.3
F= 0.3 * 8 = 2.4 N
INSTRUCTOR RESPONSE
Your error is that x should be the displacement
from equilibrium, not the full length of the rubber band.
This is easily corrected, and since you bring up
the formula for a linear restoring force we can expand on that idea a bit.
There is no reason to use F = k x here.
This relationship hasn't been introduced yet, and it's better at this point to
reason the situation out at a lower level, using fundamental definitions.
However it doesn't hurt to know how we can use
concept of a linear restoring force.
F = -k x applies if F is a linear restoring
of x, and x is measured relative to the equilibrium point.
F is regarded as the force which tends to
restore the system to its equilibrium position.
The equilibrium point is not at x = 0, which would correspond to the rubber
band being compressed to length 0.
In this case, if we assume a linear F vs. x
dependence, x = 0 occurs at the 8 cm length, where the rubber band makes the
transition from exerting no force to exerting a force.
When the rubber band is 10 cm long the
displacement from equilibrium is therefore 2 cm. The given information
tells us that the restoring force is -3 N at this point. Thus
Calculus connection:
For students who know calculus, the work
done by the rubber band between two points is the integral of this force,
with respect to position.
So in this case the work done by the rubber
band is the integral of F(x) between x = 0 and x = 2 cm:
Our antiderivative is -1.5 N / cm * x^2 / 2,
and the integral is equal to the change in the antiderivative so
The work done against the rubber band is
equal and opposite to the work done by the rubber band, and is therefore + 3
N cm.
Additional Discussion
Tension vs. Length
The rubber band exerts no tension until its
length is 8 cm. The tension is then assumed in this problem to
increase linearly, though in a real rubber band the increase would not be
perfectly linear.
In general the tension vs. length graph then
looks like this:
The tension is zero up to a certain length
(in this case 8 cm), then it begins to increase linearly. This is
indicated by the heavy red line which runs along the horizontal axis until a
certain point, then begins to increase at a constant rate.
For this specific situation the graph could
be labeled as follows, consistent with 0 tension at the 8 cm length and
tension 3 N at the 10 cm length.
The slope of the increasing part of the
graph is found from the rise and the run of the two indicated graph points
to be
The equation of the increasing part of the
graph would be
where T is the tension and L the length of
the rubber band.
Tension vs. Stretch
We could also measure tension vs. stretch,
where stretch is defined to be the length in excess of the 'zero-tension'
length.
In this case the stretch at the 8 cm length
would be zero, and the stretch at the 10 cm length would be 10 cm - 8 cm = 2
cm. Our graph of tension vs. stretch, again assuming the tension to be
a linear function of stretch, would look like the following:
In this case we could use the symbol x to
stand for the stretch, and the graph would be of T vs. x.
The slope of the graph would be the same as
before, but this time, since the graph now goes through the origin, the
function would be
where k is the slope. Since the slope
is the same as before, k = 1.5 N / cm so
Since x is the length in excess of 8 cm,
it's easy to see that x = L - 8 cm, so our two equations
T = 1.5 N / cm * (L - 8 cm)
and
T = 1.5 N / cm * x
are just two different ways of describing
the behavior of the same system.