ic_class_091014

course PHY 201

4:42 PMSunday, November 15, 2009

Class 091014

Don't remove or overwrite **** or &&&& symbols in this document. You and I both need those marks to be able to separate your answers from the question. Your work would get less scrutiny if these marks aren't both present, and it will be harder for you when you want to review it. If any of these marks are missing I will likely ask you to reinsert any missing marks and resubmit.

The Atwood machine

The Atwood machine used in class consists of a domino suspended from each side of a pulley. When we add paper clips to one side the system accelerates. The more clips, the greater the acceleration.

If you weren't in class to get data today, you may assume the following:

* for 1 added paper clip, the system moved 50 cm from rest in 6.5 swings of a 13 cm pendulum

* for 2 added paper clips, the system moved 50 cm from rest in 4.5 swings of a 16 cm pendulum

* for 3 added paper clips, the system moved 50 cm from rest in 3.5 swings of a 10 cm pendulum

* for 4 added paper clips, the system moved 50 cm from rest in 2.5 swings of a 12 cm pendulum

`q001. Give your raw data for the Atwood machine. This includes all directly observed quantities used in calculating your results.

****

10cm pendulum

`ds = 50cm

# paper clips Time (cycles)

1 4.5

2 3.5

3 3

4 3

5 2.5

6 2

7 1.75

8 1.5

&&&&

`q002. For one of your trials show in detail how you use your raw data to obtain the acceleration of the system for that trial.

****

First, find the period of the pendulum in seconds:

Period = .2 sqrt(10 cm)

= 0.632 s / cycle

Next, multiply the period by the number of half-cycles:

`dt = 4.5 cycles * 0.632 s / cycle

= 2.844 s

Use the measured displacement and the time interval to determine average velocity

v_ave = `ds / `dt

= 50cm / 2.844s

= 17.581 cm/s

We know that initial velocity is 0, and thus final velocity must be twice the average velocity:

vf = 2 * v_ave

= 2 * 17.581 cm/s

= 35.162 cm/s

We now have every quantity necessary to determine average acceleration:

a_ave = `dv / `dt

= (vf - v0) / `dt

= (35.162 cm/2) / 2.844 s

= 12.364 cm/s^2

&&&&

`q003. Give a table of acceleration vs. number of clips, one line at a time representing one trial at a time, with the number of clips then the acceleration for the trial separated by four spaces. At the beginning or the end of the table, insert another line giving the units of each column. Alternatively, you can copy a table made using a spreadsheet.

acceleration vs. number of clips:

****

# clips time(cycles) time(s) ave vel (cm/s) final vel (cm/s) ave accel (cm/s^2)

1 4.5 2.846 17.568 35.136 12.346

2 3.5 2.214 22.588 45.175 20.408

3 3 1.897 26.352 52.705 27.778

4 3 1.897 26.352 52.705 27.778

5 2.5 1.581 31.623 63.246 40.000

6 2 1.265 39.528 79.057 62.500

7 1.75 1.107 45.175 90.351 81.633

8 1.5 0.949 52.705 105.409 111.111

&&&&

`q004. Is it possible to fit a reasonable straight line to the data? How much 'leeway' do you think you have in where to fit the line?

****

My graph looks more like a curve that levels off in the middle a little bit (due to duplicate 3 cycle intervals on points 3 and 4 - probably an observational error). It is possible to try to fit a straight line to the data points, however.

&&&&

`q005. Give the coordinates of two points on the straight line you think comes as close as possible, on the average, to the points of your graph. Use one point near each end of your line, rather that two points right next to one another.

****

(1, 0), (8, 100)

&&&&

`q006. Between the two points you specified in your preceding answer, what is the rise, what is the run and what therefore is the slope? Be sure you specify the units of each of these quantities.

****

The run would be 8 paper clips. The rise would be 100 cm/s^2 in acceleration.

The slope would be 100 cm/s^2 / 8 clips = 12.5 cm/s^2 / clip

&&&&

`q007. How plausible is it that the actual acceleration vs. number of clips is in fact well represented by a straight-line graph, with the deviations of the individual points from the straight line being due mostly to experimental uncertainties?

****

Pretty plausible. My actual average deviation is 13.89 cm/s^2 / clip.

&&&&

`q008. Specify the positive direction you chose for your system. This can be specified by stating which mass goes which way in your chosen positive direction, or by stating whether the pulley rotates in the clockwise or counterclockwise direction when the system is moving in your chosen positive direction.

****

Counterclockwise

&&&&

`q009. Suppose that each side of the pulley has a mass of 20 kg, and that each paperclip has a mass of about .2 kg (these masses are not realistic for the system we observed, being much greater than the masses we used in class). Pretending that these are the actual masses in your system:

How much force is exerted by gravity on each side of the system when 1 paperclip is added to one side?

Assuming the absence of friction, what therefore is the net force on the system?

What is the mass of the system?

What therefore should be its acceleration?

****

F1 = m1 * a

= (20kg + .2kg ) * +9.8 m/s

= 20.2kg * +9.8m/s^2

= 197.96 kg*m/s^2

= 197.96 N

F2 = m2 * a

= 20kg * -9.8m/s^2

= -196 N

F_net = F1 + F2

= 197.96 N + (-196 N)

= 1.96 N

m_net = m1 + m2

= 20.2 kg + 20kg

= 40.2 kg

a = F_net / m_net

= 1.96 kg*m/s^2 / 40.2 kg

= 0.0488 m/s^2

&&&&

`q010. Find the acceleration for the system in the preceding for 3 paperclips, and for 5 paperclips, added to the same side as before.

Sketch a graph of acceleration vs. number of paperclips and fit your best straight line to the graph.

How straight do you think your line is?

What is the slope of your line? (be sure you include units)

****

3 paper clips:

F1 = m1 * a

= (20kg + .6kg ) * +9.8 m/s

= 20.6kg * +9.8m/s^2

= 201.88 kg*m/s^2

= 201.88 N

F2 = m2 * a

= 20kg * -9.8m/s^2

= -196 N

F_net = F1 + F2

= 201.88 N + (-196 N)

= 5.88 N

m_net = m1 + m2

= 20.6 kg + 20kg

= 40.6 kg

a = F_net / m_net

= 5.88 kg*m/s^2 / 40.6 kg

= 0.145 m/s^2

5 paper clips:

F1 = m1 * a

= (20kg + 1kg) * +9.8 m/s

= 21kg * +9.8m/s^2

= 205.8 kg*m/s^2

= 205.8 N

F2 = m2 * a

= 20kg * -9.8m/s^2

= -196 N

F_net = F1 + F2

= 205.8 N + (-196 N)

= 9.8 N

m_net = m1 + M2

= 21 kg + 20kg

= 41 kg

a = F_net / m_net

= 9.8 kg*m/s^2 / 41 kg

= 0.239 m/s^2

The line appears to be fairly straight, only containing 3 points. The slope of the line would be :

(0.239 m/s^2 - 0.0488 m/s^2) / (5 clips - 1 clip)

0.1902 m/s^2 / 4 clips

0.04755 m/s^2 / clip

&&&&

`q011. For this series of examples, what is the mass of a single paperclip, as a percent of the mass of the entire system?

What percent of the acceleration of gravity is the slope of the graph you made for the preceding problem?

****

The mass of a single .2kg paper clip as a percent of the mass of the entire 40.2kg system:

.2kg / 40.2kg = 0.49%

The slope of 0.04755 m/s^2 / clip as a percent of the accerlation of gravity (9.8 m/s^2):

0.04755 ms/^2 / clip / 9.8m/s^2 = 0.49%

&&&&

`q012. From the slope of the graph you made for your experiment, can you conjecture the mass of a paperclip as a percent of the mass of the system?

****

Being that the slope is 0.49% of gravity and that a paper clips is 0.49% of the mass of the system, then given the mass of the system 40.2kg, we could find the weight of the paperclip:

40.2 kg * 0.0049 = 0.19698 or approx .2kg

Your answer here is based on the example. The question asks about your experiment, in which you found that the average additional acceleration per paperclip is about 12 cm/s^2.

In light of your answer to question 11, in which you noted that the slope as a percent of the acceleration of gravity is identical to the mass of a clip as a fraction of the total mass of the system, what is your answer to the present question?

&&&&

`q013. On your graph, what is the horizontal intercept of your straight line (i.e., if the line is extended, where does the line cross the x axis)?

What are the units and the meaning of this point?

****

I would have guessed (0, 0), but looking at the graph I made in openoffice, it might be about (0, .5), which might account for the coeffecient of friction.

&&&&

`q014. If the frictional force on the system is increased, would the acceleration of the system increase, decrease or stay the same?

What effect would this have on the points of your graph?

What effect would this have on the straight line that approximates your points?

What effect would this have on the x intercept of the straight line?

****

If frictional force is increased, it would reduce the acceleration of the system.

The points on the graph would be shifted down a bit (towards the x-axis, indicating lower acceleration).

The line would remain about the same I think.

The x-intercept would occur further to the right from the origin, indicating more paper clips required to attain the same accelerations.

very good

&&&&

Energy considerations

`q015. Going back to the example problem where each mass is 20 kg and each clip has mass .2 kg, let's assume that three clips are added to the mass on the left, so that the system accelerates in the counterclockwise direction.

We want to analyze the energy situation if the system moves .7 meters in our chosen positive direction.

What downward force is exerted on a 20 kg mass by gravity?

By how much does the gravitational potential energy of the 20 kg mass on the right therefore change as the system moves +.7 meters?

Answer the same for the 20 kg mass on the left.

Answer the same for the three clips.

What therefore is the PE change of the system?

****

Force on 20kg mass by gravity:

F = m * a

= 20kg * 9.8 m/s^2

= 196 N

PE_right = 196 N * .7m

= 137.2 J

PE_left = 196 N * -.7m

= -137.2 J

PE_3_clips = 5.88 N * -.7m

= - 4.116 J

The PE change of the system is 137.2J + (-137.2 J) + (-4.116 J) = -4.116 J

&&&&

`q016. Assuming that no nonconservative forces act on the system, what therefore is its change in kinetic energy?

****

`dKE = - `dPE

= - (4.116 J

= 4.116 J

&&&&

`q017. The kinetic energy of the system is 1/2 m v^2, where m is the mass of the system.

Assuming it started from rest, its KE at the end of the interval will be equal to its change in KE.

What therefore is its KE at the end of the interval?

What is the mass of the system?

What therefore is its velocity at this point?

****

KE = 4.116 J

m = 20kg + 20kg + .6 kg = 40.6 kg

KE = 1/2 m v^2

v^2 = KE / (1/2 m)

v = sqrt( KE / .5 * m )

= sqrt(4.116 J / .5 * 40.6 kg)

= sqrt(4.116 kg * m^2/s^2 / 20.3 kg)

= sqrt(0.203 m^2 / s^2)

= 0.450 m/s

&&&&

`q018. How would your answers to the last two questions change if there is a frictional force of 3 N acting on the system?

****

The 3 N force acts opposite to the positive direction of motion, so the numbers would be a little smaller (probably cut in half).

You can be specific about this. For example you could find the work dont by the frictional force and use it to modify your estimate of `dKE.

&&&&

Vectors

`q019. The figure on the board represented three vectors, one of magnitude 10 units at 305 deg, one of magnitude 8 units at 158 deg and one of 4 units at 80 deg.

According to our estimates:

* the x and y components of the first force are respectively 45% and -86% of that force

* the x and y components of the second force are respectively -90% and -46% of that force

* the x and y components of the third force are respectively 20% and 95% of that force

Based on these estimates calculate the x and y components of the three forces.

What is the sum of all the x components?

What is the sum of all the y components?

****

Force 1:

x: 4.5 units

y: -8.6

Force 2:

x: -7.2

y: -3.68

Force 3:

x: 0.8

y: 3.8

Sum of x-components:

4.5 + (-7.2) + 0.8 = -1.9 units

Sum of y-components:

(-8.6) + (-3.68) + 3.8 = -8.48 units

&&&&

`q020. The actual percents are given by the sine and cosine functions as decimals. For example if the percents are 45% and -86%, the cosine function would give us .45 and the sine function would give us -.86.

For an angle of 305 degrees, use your calculator to find sin(305 deg) and cos(305 deg). (make sure you calculator is in 'degree' mode; using the sin/cos button find sin(305) and cos(305))

What do you get and how do the accurate values compare with our estimates?

Using the accurate values of the sine and cosine of 305 degrees, what are the x and components of the first vector?

****

sin(305) = -0.819

cos(305) = 0.574

-82% is fairly close, wihle 57% is off by 12%. Still not bad for estimates.

The new components:

x: 5.7

y: -8.2

&&&&

`q021. Use the same procedure to find the x and y components of the second and third vectors.

Give you results below, and include a brief explanation of your results.

****

Force 2:

x: sin(158) * 8 = 0.375 * 8 = 3

y: cos(158) * 8 = -0.927 * 8 = -7.416

Force 3:

x: sin(80) * 4 = 0.985 * 4 = 3.94

y: cos(80) * 4 = 0.174 * 4 = 0.696

&&&&

`q022. If you add up the x components of the three vectors, what do you get?

If you add up the y components of the three vectors, what do you get?

****

Sum of x-components:

5.7 + 3 + 3.94 = 12.64

Sum of y-components:

(-8.2) + (-7.416) + 0.696 = -14.92

I banged my head against the wall for about 30 minutes trying to figure out why these figures didn't come close to matching the estimates above and I finally figured out that the pos/neg for the estimates on the 158 degree vector were incorrect. The 158 degree vector will be in the second quadrant and will have a postive y component and a negative x component, while the given esimates were both negative.

&&&&

Homework:

Your label for this assignment:

ic_class_091014

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form. Answer the questions posed above.

You should know everything in the first six problems of Set 5 in the Introductory Problem Set, which will give you a good, and not difficult, introduction to working with vectors. A link that should get you there is at http://vhmthphy.vhcc.edu/ph1introsets/default.htm .

Very good work. Do check the appended discussion for more information and a few modifications.

Class 091014

The Atwood machine

The Atwood machine used in class consists of a domino suspended from each side of a pulley. When we add paper clips to one side the system accelerates. The more clips, the greater the acceleration.

If you weren't in class to get data today, you may assume the following:

for 1 added paper clip, the system moved 50 cm from rest in 6.5 swings of a 13 cm pendulum

for 2 added paper clips, the system moved 50 cm from rest in 4.5 swings of a 16 cm pendulum

for 3 added paper clips, the system moved 50 cm from rest in 3.5 swings of a 10 cm pendulum

for 4 added paper clips, the system moved 50 cm from rest in 2.5 swings of a 12 cm pendulum

`q001. Give your raw data for the Atwood machine. This includes all directly observed quantities used in calculating your results.

****

Using the information to be assumed by students who weren't present:

for 1 added paper clip, the system moved 50 cm from rest in 6.5 swings of a 13 cm pendulum

for 2 added paper clips, the system moved 50 cm from rest in 4.5 swings of a 16 cm pendulum

for 3 added paper clips, the system moved 50 cm from rest in 3.5 swings of a 10 cm pendulum

for 4 added paper clips, the system moved 50 cm from rest in 2.5 swings of a 12 cm pendulum

&&&&

`q002. For one of your trials show in detail how you use your raw data to obtain the acceleration of the system for that trial.

****

Using the data for 2 added paper clips:

The system moved 50 cm in 3.5 swings of a 16 cm pendulum.

A 16 cm pendulum has period .2 sqrt(16) sec = .8 sec.

3.5 swings therefore covers 4.5 * .8 sec = 3.6 sec.

average velocity is therefore 50 cm / (3.6 sec) = 15 cm/s, approx.

initial velocity is 0 so assuming uniform acceleration (linear v vs. t) final velocity is 30 cm/s

acceleration is therefore 30 cm/s / (3.6 s) = 8 cm/s^2, approx..

(alternatively the equations of uniformly accelerated motion, with v0 = 0, `ds = 50 cm and `dt = 2.8 s, give the same result)

&&&&

acceleration vs. number of clips:

****

1, 4.6

2, 8

3, 20

4, 33

The number of clips is just a pure number, while the accelerations are in cm/s^2.

&&&&

`q004. Is it possible to fit a reasonable straight line to the data? How much 'leeway' do you think you have in where to fit the line?

****

The second, third and fourth points lie along a nearly straight line. The first point distorts the picture significantly, pulling the best-fit straight line away from the line of the last three points in such a way as to raise its y intercept and reduce its slope.

A reasonable straight-line approximation for this particular data might be close to the best-fit line, which is near y = 10 x - 8. However the fact that the points are not tightly grouped about this straight line indicates a fair amount of uncertainty in both slope and intercept.

The rise of the graph is the change in acceleration, in cm/s^2.

The run is the number of clips.

So the slope is (change in acceleration) / (change in number of clips) = rate of change of acceleration with respect to number of clips.

This is interpreted as the additional acceleration we expect for each added paper clip.

For the given data the slope is 10; we therefore expect an additional 10 cm/s^2 of acceleration for each added clip.

However recall the previous note about the uncertainty in the acceleration.

&&&&

****

A reasonable straight line for the data used here might pass through the points (1 clip, 0 cm/s^2) and (5 clips, 40 cm/s^2). Neither is a data point.

&&&&

`q006. Between the two points you specified in your preceding answer, what is the rise, what is the run and what therefore is the slope? Be sure you specify the units of each of these quantities.

****

(1 clip, 0 cm/s^2) and (5 clips, 40 cm/s^2) is (40 cm/s^2 - 0 cm/s^2) / (5 clips - 1 clip) = 10 (cm/s^2) / clip.

&&&&

****

The fact that the points are cannot be tightly grouped about a single straight line indicates a fair amount of uncertainty in both slope and intercept. However the dispersion about the straight line is not extreme, so the slope is still expected to be somewhat accurate.

&&&&

****

We will assume here that the greater mass is on the right, and the positive direction will be the direction in which the mass on this side descends while the mass on the left ascends.

The pulley will in this case rotate in the clockwise direction.

(note for later reference that the positive direction for rotation is generally regarded as the counterclockwise directions, for reasons you will soon see)

&&&&

How much force is exerted by gravity on each side of the system when 1 paperclip is added to one side?

Assuming the absence of friction, what therefore is the net force on the system?

What is the mass of the system?

What therefore should be its acceleration?

****

The side with the 20 kg mass will experience a downward gravitational force of 20 kg * 9.8 m/s^2 = 196 Newtons.

The side with the paperclip will have a mass of 20.2 kg and will experience a downward gravitational force of 20.2 kg * 9.8 m/s^2 = 198 Newtons.

The 196 Newton force is in the direction previously chosen as negative, the 198 Newton force in the direction chosen as positive.

The net force is therefore

F_net = 198 N + (-196 N) = 2 N.

The system consists of two 20 kg masses and the .2 kg clip, all accelerating together. So the mass of the system is 40.2 kg.

Its acceleration is therefore

a = F_net / m = 2 N / (40.2 kg) = .050 m/s^2.

&&&&

`q010. Find the acceleration for the system in the preceding for 3 paperclips, and for 5 paperclips, added to the same side as before.

Sketch a graph of acceleration vs. number of paperclips and fit your best straight line to the graph.

How straight do you think your line is?

What is the slope of your line? (be sure you include units)

****

The systems will have net forces of 6 N and 10 N, respectively, and masses of 40.6 kg and 41 kg, respectively. So the accelerations will be

a = F_net / m = 6 N / (40.6 kg) = .148 m/s^2 and

a = F_net / m = 10 N / (41 kg) = .244 m/s^2

Thus we have acceleration vs. number of clips, according to the following table:

1, .048

3, .148

5, .244

with the first column representing number of clips, the second accel in m/s^2.

A straight line fits these points very well, though not exactly.

The slope of the line is very close to .050 m/s^2.

The interpretation of the slope is that the acceleration changes by 050 (m/s^2) for each additional clip. So the slope could be expressed as .050 m/s^2 / clip.

&&&&

`q011. For this series of examples, what is the mass of a single paperclip, as a percent of the mass of the entire system?

What percent of the acceleration of gravity is the slope of the graph you made for the preceding problem?

****

The mass of the clip is .2 kg, the mass of the entire system varies from 40.2 kg to 41 kg.

The mass of the clip as a proportion of the mass of the system is between .2 / 40.2 = .00498 and .2 / 41 = .00493; so the mass lies between .0050 and .0049 of the mass of the system.

In terms of percents, the mass of a clip is between .5% and .49% of the mass of the system.

The acceleration of gravity is 9.8 m/s^2. The slope of the line is about .050 m/s^2. This is .050 m/s^2 / (9.8 m/s^2) = .0051, or .51%.

It is worth noting that this result is very close to the .49% or .50% we obtained for the mass of a clip as a percent of the mass of the system.

So, for this example at least, the mass of a clip is about .5% the mass of the system, and the slope of the graph is about .5% the acceleration of gravity.

Remembering the the slope of the graph represents the additional acceleration expected from the addition of 1 clip, the similarity in these results should be striking.

&&&&

`q012. From the slope of the graph you made for your experiment, can you conjecture the mass of a paperclip as a percent of the mass of the system?

****

The additional acceleration corresponding to 1 single paperclip is .5% the acceleration of gravity.

If gravity acted unimpeded on the mass of the entire system, then the net force on the system would be its weight, and it would acceleration with 100% the acceleration of gravity.

If the net force on the system was half the force exerted by gravity (i.e., if the net force was equal to half its weight), it would acceleration with half that acceleration.

To accelerate with 10% the acceleration of gravity, the net force would have to be 10% of weight of the system.

If the system accelerates with 2% the acceleration of gravity, then the net force must be 2% of the weight of the system.

Based on the graph the system accelerates with .50% the acceleration of gravity.

So according to the graph, the additional mass of a single paperclip would corresponds to .50% the mass of the system.

&&&&

`q013. On your graph, what is the horizontal intercept of your straight line (i.e., if the line is extended, where does the line cross the x axis)?

What are the units and the meaning of this point?

****

Using preceding results, the horizontal intercept was at (1, 0), indicating that a mass difference equivalent to 1 paperclip corresponded to 0 acceleration.

To the left of the intercept our straight line indicates a negative acceleration. So a mass difference equivalent to less than 1 paperclip would correspond to a negative acceleration. We did not make any observations in this experiment related to negative acceleration.

To the right of the intercept our graph is positive. So a mass difference equivalent to more than 1 paperclip would correspond to a positive acceleration. This is consistent with our observations in this experiment.

&&&&

`q014. If the frictional force on the system is increased, would the acceleration of the system increase, decrease or stay the same?

What effect would this have on the points of your graph?

What effect would this have on the straight line that approximates your points?

What effect would this have on the x intercept of the straight line?

****

More friction would 'hold the system back' more. This would have two effects on the system:

It would require more clips on the 'positive' side before the system would accelerate in the positive direction.

A given number of clips would result in less acceleration that if friction was absent or negligible.

The corresponding effects on the graph will be as follows:

The acceleration for any given number of clips will be reduced as a result of friction. Since acceleration is graphed as the 'vertical' quantity, this will lower each point of the graph. The straight-line approximation will therefore be lowered as well.

The x intercept, which corresponds to the number of additional clips required to begin accelerating the system in the positive direction, would increase due to friction (i.e., the x intercept of the graph would be further to the right).

If the force exerted by friction is the same regardless of velocity, then the frictional force will reduce the acceleration for each trial by the same amount. This will have the effect of lowering each point of the graph by the same amount. This will affect the position, but not the slope, of the straight-line approximation to the graph.

It is well worth noting that if the frictional force is constant (e.g., if it is independent of position and velocity), the graph will be lower than the 'ideal' no-friction graph, but the slope of the graph will not be affected.

&&&&

Energy considerations

We want to analyze the energy situation if the system moves .7 meters in our chosen positive direction.

What downward force is exerted on a 20 kg mass by gravity?

By how much does the gravitational potential energy of the 20 kg mass on the right therefore change as the system moves +.7 meters?

Answer the same for the 20 kg mass on the left.

Answer the same for the three clips.

What therefore is the PE change of the system?

****

Gravity exerts a force of 20 kg * 9.8 m/s^2 = 196 N on a 20 kg mass, and a force of 2 N on a .2 kg mass.

As the mass on the right descends, gravity acts downward, so the gravitational force on the right-hand mass is in the same direction as its displacement. Gravity therefore does positive work, equal to 196 N * .7 m = 137 Joules, on the left-hand mass. The PE of this mass, being equal and opposite to the work done by the conservative force acting on it, therefore decreases by 137 J. For this mass we say that `dPE = -137 J (if we prefer to use different symbols for the PE change of each component, we might distinguish this one by stating that `dPE_rh_mass = -137 J).

As the mass on the left ascends, gravity acts downward, so the gravitational force on the left -hand mass is in the direction opposite its displacement. Gravity therefore does negative work, equal to -196 N * .7 m = -137 Joules, on the left-hand mass. The PE of this mass, being equal and opposite to the work done by the conservative force acting on it, therefore increases by 137 J. For this mass we say that `dPE = +137 J (if we wish to distinguish our symbols we might say that `dPE_lh_mass = +137 J).

The three clips have total mass .6 kg. They are on the right-hand side of the system. The analysis is similar to that of the first paragraph, and we conclude that the PE of the clips decreases by .6 kg * 9.8 m/s^2 * .7 m = 4.1 Joules. For the mass of the clips we say that `dPE = -4.1 Joules (or to distinguish symbols, `dPE_clips = -4.1 J)..

The PE change of the system is equal to that of its components:

`dPE_system = -137 J + 137 J - 4.1 J = -4.1 J.

(if we prefer to distinguish our symbols we might write

`dPE_system = `dPE_rh_mass + `dPE_lh_mass + `dPE_clips = -137 J + 137 J - 4.1 J = -4.1 J).

&&&&

`q016. Assuming that no nonconservative forces act on the system, what therefore is its change in kinetic energy?

****

If no nonconservative forces act on the system then `dW_NC_ON = 0 so that our energy conservation law

`dW_NC_ON = `dKE + `dPE implies that

`dKE + `dPE = 0

we conclude that

`dKE = - `dPE = - (-4.1 J) = +4.1 J.

&&&&

`q017. The kinetic energy of the system is 1/2 m v^2, where m is the mass of the system.

Assuming it started from rest, its KE at the end of the interval will be equal to its change in KE.

What therefore is its KE at the end of the interval?

What is the mass of the system?

What therefore is its velocity at this point?

****

Since, starting from rest, its KE at the end of the interval is equal to `dKE we see that at the end of the interval

KE = 4.1 J.

The mass of the system with 3 clips is 40.6 N.

Since by the definition of kinetic energy we have

KE = 1/2 m v^2

we can find the velocity of the system by solving this equation for v.

Multiplying both sides of the equation by 2 / m we have

2 / m * KE = 2 / m * (1/2 m v^2) so that

2 KE / m = v^2 and

v = +- sqrt( 2 KE / m).

Substituting 4.1 J for the KE and 40.6 kg for m we obtain

v = +- sqrt( 2 * 4.1 J / (40.6 kg) ) = +- .14 m/s, approx..

Our velocity could therefore be either +.14 m/s or -.14 m/s. The choice of which depends on our original choice of positive direction.

In this case we specified the positive direction to be that in which the greater mass descends. Since for this particular trial it is the greater mass that descends, we conclude that its final velocity is

v = +.14 m/s.

&&&&

`q018. How would your answers to the last two questions change if there is a frictional force of 3 N acting on the system?

****

A 3 Newton frictional force would act in the direction opposite motion, and would therefore do negative work. The frictional force is nonconservative. Assuming no additional forces act on the system, we would therefore have

`dW_NC_ON = - (3 N) * (.7 m) = -2.1 Joules.

Our energy conservation equation would therefore be

`dW_NC_ON = `dKE + `dPE

-2.1 J = `dKE + (-4.1 J) so that

`dKE = -2.1 J + 4.1 J = 2 J.

The velocity of the system would therefore be

v = +- sqrt( 2 KE / m) = +- sqrt( 2 * 2 J / (40.6 kg) ) = +- .10 m/s.

Again our choice of positive direction implies the positive solution so we have final velocity

v = +.10 m/s.

&&&&

Vectors

`q019. The figure on the board represented three vectors, one of magnitude 10 units at 305 deg, one of magnitude 8 units at 158 deg and one of 4 units at 80 deg.

According to our estimates:

the x and y components of the first force are respectively 45% and -86% of that force

the x and y components of the second force are respectively -90% and 46% of that force

the x and y components of the third force are respectively 20% and 95% of that force

Based on these estimates calculate the x and y components of the three forces.

What is the sum of all the x components?

What is the sum of all the y components?

****

Let's call the three forces F_1, F_2 and F_3.

The x and y components of the three forces are

F_1x = .45 * 10 units = 4.5 units and F_1y = -.86 * 10 units = -8.6 units

F_2x = -.90 * 8 units = -7.2 units and F_2y = -.46 * 8 units = -3.7 units

F_3x = .20 * 4 units = .8 units and F_2y = .95 * 4 units = 3.8 units

Using F_tot for the total force, the sum of the x and y components components are

F_tot_x = F_1x + F_2x + F_3x = 4.5 unit + (-7.2 units) + 0.8 units = -1.9 units

F_tot_y = F_1y + F_2y + F_3y = -8.6 units + (- 3.7 units ) + 3.8 units = -8.5 units.

&&&&

For an angle of 305 degrees, use your calculator to find sin(305 deg) and cos(305 deg). (make sure you calculator is in 'degree' mode; using the sin/cos button find sin(305) and cos(305))

What do you get and how do the accurate values compare with our estimates?

Using the accurate values of the sine and cosine of 305 degrees, what are the x and components of the first vector?

****

With a calculator in degree mode, as opposed to radian mode, you get

cos(305 deg) = 0.5735764363 (compare with our estimate .45)

sin(305 deg) = -0.8191520442 (compare with our estimate -.86)

These results are reasonably close to those of our ballpark estimate.

NOTE: If you got -.26 and -.97, corresponding to -26% and -97%, you didn't have your calculator in degree mode. Those results are what you would get if your calculator was in radian mode.

Using 2-significant-figure approximations the x and y components of the first vector are

F_1x = 10 units * cos(305 deg) = 10 units * .57 = 5.7 and

F_1y = 10 units * sin(305 deg) = 10 units * (-.82) = -8.2.

&&&&

`q021. Use the same procedure to find the x and y components of the second and third vectors.

Give you results below, and include a brief explanation of your results.

****

Results for the sines and cosines are:

cos(158 deg) = -0.9271838545 (compare with our estimate -.90)

sin(158 deg) = 0.3746065934 (compare with our estimate .46)

cos(90 deg) = 0.1736481776 (compare with our estimate .20)

sin(80 deg) = 0.9848077530 (compare with our estimate .95)

F_2x = 8 units * (-.93) = -7.4 units and F_2y = 8 units * .46 = 3.6 units

F_3x = 4 units * .17 = 0.7 units and F_3y = 4 units * .98 = 3.9 units.

&&&&

`q022. If you add up the x components of the three vectors, what do you get?

If you add up the y components of the three vectors, what do you get?

****

The x and y components of the three vectors are:

F_1x = 10 units * .57 = 5.7 units and F_1y = 10 units * (-.82) = -8.2 units

F_2x = 8 units * (-.93) = -7.4 units and F_2y = 8 units * .46 = 3.6 units

F_3x = 4 units * .17 = 0.7 units and F_3y = 4 units * .98 = 3.9 units.

Thus

F_x = F_1x + F_2x + F_3x = 5.7 units + (-7.4 units) + 0.7 units = -.10 units and

F_y = F_1y + F_2y + F_3y = -8.2 units + 3.6 units + 3.9 units = -.7 units.

Homework:

Your label for this assignment:

ic_class_091014

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form. Answer the questions posed above.