#$&*
course phy 201
11/13 9:30 pm
`q001. A stretched rubber band is used to set a system in motion. The system comes to rest. For each system described below identify the conservative forces and nonconservative forces acting on the system, the directions of these forces, the direction of the displacement through which each force acts, the work done by each of these forces and whether that work is positive or negative, and state how the kinetic and potential energies change as a result.
System 1: A rotating strap loaded at various points with weights, resting on a domino.
****
Conservative: rubber band on strap;
Nonconservative: friction, air resistance
Direction of displacement: circular motion.
@&
The force acts in the direction of the rubber band. That direction will change slightly as the rubber band's length returns to its original value.
*@
Work done by conservative forces: positive
Work done by nonconservative forces: negative;
Kinetic energy will increase, potential energy will decrease;
#$&*
System 2: A wood block on the tabletop, across which it slides.
****
Conservative: force of gravity down on block, force of rubber band on block,
Nonconservative: friction, force of tabletop returning gravity’s push
Direction of displacement: Direction of the applied rubber band
Work done by conservative forces: positive
Work done by nonconservative forces: negative
Kinetic energy will increase, potential energy will decrease
#$&*
System 3: An arrow which is shot straight up.
****
conservative forces: force of rubber band on arrow, force of gravity;
nonconservative forces: air resistance
direction of displacement: upward into the air
Work done by conservative forces: rubber band: positive, gravity: negative
Work done by nonconservative forces: negative;
Kinetic energy will decrease, potential energy will increase.
#$&*
System 4: A rotating strap loaded at various points with weights and a two sheets of tissue paper each weighted down at the bottom and attached near the ends of the strap, resting on a domino.
****
conservative forces: force of rubber band on strap.
Nonconservative forces: friction between strap and domino, friction between weighted tissues and resting surface
Direction of displacement: circular movement, rotation based on the starting angle of the rubber band.
Work done by conservative forces: positive
Work done by nonconservative forces: negative
KE increases, PE decreases
#$&*
System 5: A toy car released on a slightly inclined tabletop.
****
conservative forces: force of gravity, force of rubber on car
nonconservative forces: return force by incline, friction.
Direction of displacement: down the slope
Work done by conservative forces: positive
Work done by non conservative forces: negative
KE increases, PE decreases
#$&*
System 6: A pulley with unbalanced weights supported by a string over both sides.
****
Assuming rubber band is set on lesser weight.
Conservative forces: Gravity, force of rubber band
Nonconservative forces: friction
Direction of displacement: Directly down by the lesser weight until rest
Work done by conservative forces: Gravity does overall negative work, though positive work is done since it effects both sides. Rubber band, positive work
Work done by nonconservative forces: negative
KE decreases, PE increases
#$&*
`q002. A rubber band chain experiences tensions of 0 N, 1 N and 1.6 N when its 'stretch' is respectively 0, 5 cm and 10 cm.
What is your best estimate of the work done to stretch in the chain from 0 'stretch' to the 10 cm 'stretch'?
****
average force: 0.8 newtons; displacement: 10 cm, or 0.1 meters; work done: 0.08 joules;
@&
This isn't the best possible estimate. It ignores the fact that at the 5 cm length the force is 1 Newton. Your solution simply assumes a linear force relationship, when the data clearly indicates that the relationship is not linear.
*@
@&
Given only the first and last data points, your result would be the best reasonable estimate.
*@
#$&*
Upon release the rubber band accelerates a rotating system from rest. The rotating system consists of a thin strap whose mass we will regard as negligible and whose length is 20 cm, rotating about its center with two 50-gram magnets on each end. What is its moment of inertia?
****
I = 1/12 m * l^2; I = 1/12 100 g * 20 cm^2; I = 8.33g * 400 cm^2 I = 3333.33 g cm^2;
@&
The system also includes the two 50-gram magnets, whose moments of inertia must be added to that of the strap.
*@
#$&*
When the rubber band has separated from the system its angular velocity is 10 radians / second (about 570 degrees / second). What is its kinetic energy?
****
KE = ½ I * omega^2; KE = 1666.67 g cm^2 * 100 radians/s^2; KE = 166,667 g cm^2/s^2; 1000 grams per kg; 100 cm per meter; 1.667 joules
#$&*
What percent of the work required to stretch the rubber band is transferred to the kinetic energy of the system?
****
roughly 200%
#$&*
How far does would a magnet travel in one 360-degree revolution, if it wasn't slowing down?
****
KE = ½ m * v^2; 1.667 joules = 0.05 kg * v^2; 33.34 m^2/s^2 = v^2; 5.77 m/s = v; angular velocity: 560 degrees/second; time to reach 360 degrees: 0.643 seconds; 5.77 m/s * 0.643 seconds = 3.71 meters;
#$&*
How far would it travel in a second?
****
5.77 meters;
#$&*
What therefore is its speed?
****
5.77 m/s
#$&*
What is its kinetic energy at that speed?
****
1.667 Joules
#$&*
If instead of transferring its energy to the rotating system the rubber band was used to 'shoot' the two magnets (now separated from the strap) straight upward, how much gravitational PE would they gain before coming to rest, and how high would they be at that point?
****
energy applied: 1.667 joules; I am not sure how to proceed after this point
@&
If the magnet's height changes by `ds_y, by how much does its gravitational PE change?
For what height change `ds_y would this change be equal to the 1.67 Joules of energy provided to the system?
*@
#$&*
If the rubber band was used to 'shoot' the magnets across the table, where they slide against a frictional force equal to 30% of their weight, how far would they travel?
****
weight: 0.1 kg; 9.8 * 0.1 = 0.98 newtons; force applied by rubber band: 0.8 newtons; friction: 0.294; initial acceleration of magnets: 8 m/s^2; opposing acceleration: 2.94 m/s^2; I’m pretty sure at this point I’ve screwed up what I’m supposed to be doing, because I have no idea what to do at this point.
@&
The frictional force is .294 Newtons. Through what displacement would this force dissipate the 1.67 Joules of energy?
*@
#$&*
`q003. An 'ideal' rubber band has a force vs. stretch graph which passes through the origin and has slope .8 Newtons / centimeter.
What is its tension when its stretch is 10 cm?
****
8 newtons;
#$&*
How much work does it take to stretch it from position 0 cm to position 10 cm?
****
average force: 4 newtons; work: 40 joules;
@&
4 Newtons * 10 cm is not 40 Joules, it's .4 Joules.
Be careful to do complete units analysis.
*@
#$&*
In terms of the symbol x, what is its tension when its stretch is x?
****
0.8 * x cm;
#$&*
In terms of x, how much work does it take to stretch it from the 0 cm position to position x?
****
1/2(0.8 * x) * x
#$&*
What is its elastic potential energy at the 10 cm position, and at position x?
****
at 10 cm: 40 joules;
at position x: ½(0.8 * x) * x
@&
Very good. Your expression could be written
PE = 1/2 * (0.8 N / cm) * x^2.
0.8 N / cm is called the force constant.
In general if k is the force constant then the potential energy is 1/2 k x^2.
*@
#$&*
`q004. A force is directed along a line in the xy plane. Between two points of the line the 'rise' is 15 cm and the 'run' is 20 cm.
The line segment between these points is the hypotenuse of a right triangle whose legs are the 'rise' and the 'run'. What is the length of that segment?
****
25 cm
#$&*
What percent of that length is the 'rise'?
****
60%
#$&*
What percent of that length is the 'run'?
****
80%
#$&*
Suppose the force is 50 Newtons.
****
#$&*
If the y component of that force is in the same proportion to the force as the 'rise' to the hypotenuse, what is the y component?
****
30 newtons
#$&*
If the x component of that force is in the same proportion to the force as the 'rise' to the hypotenuse, what is the x component?
****
40 newtons
#$&*
Estimate the angle between the 'run' leg of the triangle and the hypotenuse.
****
adjacent/hypotenuse = cos; arcos(20/25) = 0.6435 radians; roughly 36.88 degrees;
#$&*
"
&#Good responses. See my notes and let me know if you have questions. &#