`q001. The work done on a certain rubber band, as it stretches from length 64 cm to 68 cm, is represented by the area of a force vs. length trapezoid with 'graph altitudes' 0.4 Newtons and 0.8 Newtons, and trapezoid 'width' 4 cm. What is the area of this trapezoid?

A = 1/2b*h

A = .4 N + .8 N/2 * 4

A= 2.4 cm

#$&*

What is the average force exerted over this 4 cm interval?

0.4 N + 0.8 N / 2 = 0.6 N

#$&*

What is the product of the average force and the displacement?

0.6N * 4 cm = 2.4 Ncm

#$&*

Why is the answer to the last question equal to the area of the trapezoid?

Because the 4 cm represents the displacement and the average force represents the .5 b in the first equation.

You need to answer this in terms of the trapezoid and its interpretation.

#$&*

How much work is done on the rubber band?

W = f*d

W = 0.4 N* 4cm

W = 1.6 Ncm

`dW = F_ave * `ds. `ds is 4 cm but F_ave is not 0.4 N.

#$&*

Where did the energy to do that work come from?

The energy came from potential energy that was stored in the stretched rubber bands just waiting to be turned into kinetic energy.

#$&*

`q002. If the rubber band in the preceding exerts a force of 0 Newtons when its length is 60 cm, then assuming that the force vs. length graph is a straight line what is the area of the corresponding trapezoid?

a = (0N + .8N/2) (8cm)

a = 3.2 cm

area = (0N + .8N)/2 (8cm) = 3.2 N * cm, not 3.2 cm. Note also the sign of grouping.

#$&*

What is the average force?

Fave = 0+.8 / 2

Fave = 0.4

#$&*

Why should the work done on the rubber band for this interval be equal to the area under the trapezoid?

Because displacement * Fave is equal to the amount of work done on the rubber band. 0.4N *8 cm = 3.2 cm which is the area under the trapezoid.

the unit of the result is N * cm, not cm.

#$&*

`q003. The kinetic energy of a mass as it moves from point to point is 1/2 m v^2, where m is its mass and v its velocity.

A typical domino has mass around 0.015 kilograms.

When a domino hits the floor after being dropped from a height of 1 meter its velocity is about 4.4 meters / second.

What is its kinetic energy at that velocity?

KE = ½ MV^2

KE = ½ (0.015kg)(4.4m/s)^2

KE = 0.1452 J

#$&*

Gravity exerts a force of about 0.15 Newtons on this domino. How much work is done on it by gravity as the domino falls that 1 meter to the floor?

W = Fave * displacement

W = 0.15 N (9.68m)

W = 1.452 J

The displacement is 1 meter, not 9.68 meters.

#$&*

A Newton is a kilogram * meter / second^2. How does the kinetic energy gained by the domino as it falls compare to the work done on it by gravity?

There is more work put into making the domino fall than is required of the kinetic energy that is already in motion. The two values are off by one decimal point but they contain the same values.

If you correct the displacement the two are the same, except for a small roundoff error but to some discrepancy related to the given 4.4 m/s velocity.

#$&*

`q004. Give your data from today's experiments, along with a brief explanation.

Given in (time in s, distance in cm)

(1.6,11), (1.8,13.5), (1.6,14.5), (2.0,26), (2.3,34.5), (3.3,89), (3.8,1.3)

Your description is insufficient. You won't be able to look at this in 6 months and tell what was done. I'm not even completely sure what was being timed here, though I suspect it's the toy car on the level tabletop.

#$&*

Sketch a graph of coasting distance vs. proximity of the magnets and give a brief description of the graph.

The graph has a negative slope and the points could easily make a line of best fit. The graph shows that there is constant acceleration.

You don't appear to have given any data related to coasting distance vs. proximity.

#$&*

What is your best result for the acceleration of the car as it moves across the tabletop?

The best result for acceleration was the car that traveled 89 cm in 3.3 s and had a V of 26.97 cm/s and an acceleration of 8.17 cm/s^2.

when analyzing motion on an interval there is no such quantity as v. There is v0, vf, vAve, and `dv, but no v. Which of these quantities are you referring to? (no need to answer; it's clear that this is vAve. The point is that you need to remain sure of the meaning of each quantity, and calling this quantity v is simply not specific enough).

The result of being imprecise about the meaning of that quantity is that you divided it by `dt and called the result acceleration. However acceleration is not vAve / `dt.

#$&*

Let's assume that the object consisting of the car plus the magnet and domino(es) has a mass of 0.100 kilograms. How much force does it take to accelerate it at the rate you just gave?

F= ma

F = 0.1kg * 0.0817 m/s^2

F = 0.00817 N

#$&*

How much work is therefore done by the accelerating force as the car travels 10 centimeters?

W = force * displacement

W = 0.00817 N * 0.1 m

W = 0.000817 J

this answer and the preceding need to be revised, after you revise your calculation of acceleration.

#$&*

`q005. We will define rubber band stretch as its length, minus the shortest length observed. So for example if you observed chain lengths 21.5, 23.0, 24.0 and 24.5 centimeters, you would subtract the 21.5 centimeters to get stretches of 0, 1.5, 2.5 and 3.0 centimeters.

Sketch a graph of rubber band stretch vs. magnet proximity, including an approximate curve following the trend of your data, and briefly describe your graph.

The graph has a negative slope and the points can form a line of best fit easily.

#$&*

Estimate the area under your graph to the right of length 4 centimeters.

area =( 0.5 cm +1.17 cm) / 2 * (3.5)

area = 2.9225 cm

it's not possible for me to evaluate this without your data

#$&*

Estimate the area under your graph to the right of length 3 centimeters.

area = (.5 cm+1.65 cm)/2 * (4.5)

area = 1.865 cm

#$&*

If every centimeter of stretch corresponds to a force of 0.1 Newton, how many Newton * cm are represented by each of your graph areas?

2.9225 cm * 0.1N = .29225 Ncm

1.865 cm * 0.1N= 0.1865 Ncm

These results are plausible, but need to be viewed in terms of your data.

#$&*

`q006. University Physics:

The graph of coasting distance vs. proximity is given by y = 7800 cm^3.32 * x^(-2.32), where y is coasting distance and x is proximity.

Coasting distance is assumed to be proportional to the work done by the magnetic forces. If A is the conversion factor from centimeters of coasting to Newton * cm of work, then, W = 7800 cm^3.32 * A x^(-2.32) gives the work done by the magnetic force as a function of proximity.

Evaluate your function to find the force at x = 3, 5 and 7 cm.

****

#$&*

What is the average rate of change of W with respect to x on each of the two x intervals corresponding to your previous results?

****

#$&*

What is the meaning of this rate of change? (hint, if needed: what do you get if you multiply this rate of change on an interval by `dx?)

****

#$&*

What therefore is the relationship between the work vs. x function and the force vs. x function?

****

#$&*

Based on W = 7800 cm^3.32 * A x^(-2.32), what is the function for force vs. x?

****

#$&*

The conversion factor A is the number of Newton * cm of work done by the accelerating force, per centimeter of coasting. Assuming (as in an earlier question) that your car plus magnets and dominoes has a mass of .1 kilogram, you can use the acceleration of your car to figure out the frictional force which brings it to rest. Then you can figure out how many Newton * cm of work are done as the car moves 1 centimeter. This result will be your value of A.

So what is the value of A?

****

#$&*

What therefore is the force when proximity x is 4 centimeters?

****

#$&*

According to your force vs. stretch graph, if each centimeter of stretch corresponds to .16 Newtons of force, how much force is exerted when proximity x is 4 centimeters?

****

#$&*

Your two results for force should theoretically be the same. However experimental uncertainties, not to mention the law of natural cussedness than infests actual physics systems causing them to misbehave in the most inconvenient possible manner, dictate that there will be a difference. How well do your two results compare and to what do you attribute any differences?

****

#$&*

@&
You're doing well, but there are some things that need revision.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.

9-24-12 classwork

9-24-12 classwork

You need to answer this in terms of the trapezoid and its interpretation.

`dW = F_ave * `ds. `ds is 4 cm but F_ave is not 0.4 N.

area = (0N + .8N)/2 (8cm) = 3.2 N * cm, not 3.2 cm. Note also the sign of grouping.

the unit of the result is N * cm, not cm.

The displacement is 1 meter, not 9.68 meters.

If you correct the displacement the two are the same, except for a small roundoff error but to some discrepancy related to the given 4.4 m/s velocity.

Your description is insufficient. You won't be able to look at this in 6 months and tell what was done. I'm not even completely sure what was being timed here, though I suspect it's the toy car on the level tabletop.

You don't appear to have given any data related to coasting distance vs. proximity.

The result of being imprecise about the meaning of that quantity is that you divided it by `dt and called the result acceleration. However acceleration is not vAve / `dt.

this answer and the preceding need to be revised, after you revise your calculation of acceleration.

it's not possible for me to evaluate this without your data

These results are plausible, but need to be viewed in terms of your data.

@&
You're doing well, but there are some things that need revision.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.

9-24-12 classwork

9-24-12 classwork

You need to answer this in terms of the trapezoid and its interpretation.

`dW = F_ave * `ds. `ds is 4 cm but F_ave is not 0.4 N.

area = (0N + .8N)/2 (8cm) = 3.2 N * cm, not 3.2 cm. Note also the sign of grouping.

the unit of the result is N * cm, not cm.

The displacement is 1 meter, not 9.68 meters.

If you correct the displacement the two are the same, except for a small roundoff error but to some discrepancy related to the given 4.4 m/s velocity.

Your description is insufficient. You won't be able to look at this in 6 months and tell what was done. I'm not even completely sure what was being timed here, though I suspect it's the toy car on the level tabletop.

You don't appear to have given any data related to coasting distance vs. proximity.

The result of being imprecise about the meaning of that quantity is that you divided it by `dt and called the result acceleration. However acceleration is not vAve / `dt.

this answer and the preceding need to be revised, after you revise your calculation of acceleration.

it's not possible for me to evaluate this without your data

These results are plausible, but need to be viewed in terms of your data.

@& You're doing well, but there are some things that need revision.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).Be sure to include the entire document, including my notes.

@&
You're doing well, but there are some things that need revision.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.