cq_1_121

#$&*

Phy 201

Your 'cq_1_12.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** **

Copy the problem below into a text editor or word processor.

• This form accepts only text so a text editor such as Notepad is fine.

• You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.

• If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.

• As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.

You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.

________________________________________

Masses of 5 kg and 6 kg are suspended from opposite sides of a light frictionless pulley and are released.

• What will be the net force on the 2-mass system and what will be the magnitude and direction of its acceleration?

answer/question/discussion: ->->->->->->->->->->->-> :

F1 = -9.8 m/s^2 upward

F2 = 9.8 m/s^2 downward

F1 = 5 * -9.8 m/s^2

F2 = 6 * 9.8 m/s^2

F1 = -5 * 9.8 m/s^2 = -49 Newtons

F2 = 6 * 9.8 m/s^2 = 58.8 Newtons

W2 + (-T) – W1 + (-T)

Fnet = F2 – F1

58.8 – 49 = or = 11A = 9.8 Force Net

A = 9.8 / 11 = 0.9 m/s^2

Upward direction at 5 kg, Downward direction at 6 kg

A

#$&*

• If you give the system a push so that at the instant of release the 5 kg object is descending at 1.8 meters / second, what will be the speed and direction of motion of the 5 kg mass 1 second later?

answer/question/discussion: ->->->->->->->->->->->-> :

v0 = 1.8 m/s

v01 = -1.8 m/s

v(t) = v01 + at

a = 0.9, t = 1

v(t) = -0.9 m/s = velocity in one second

downward

#$&*

• During the first second, are the velocity and acceleration of the system in the same direction or in opposite directions, and does the system slow down or speed up?

answer/question/discussion: ->->->->->->->->->->->-> :

As initial velocity of the system is with 5 down and 6 up, velocity is -0.9

that would be acceleration, not velocity

, still downward, but with less speed, the system is slowing down eventually moving in the other direction.

#$&*

** **

1.5 hours

** **

10/4 5 pm

&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.

Then please compare your old and new solutions with the expanded discussion at the link

Solution

Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.

If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#