asst 8

course Phy 121

HK}ɷyDRassignment #001

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Liberal Arts Mathematics I

06-25-2007

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assignment #008

008. `query 8

Physics I

06-25-2007

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11:08:56

In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?

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RESPONSE -->

constant accel * the rate at which 'ds is reached is proportional to velocity

confidence assessment: 1

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11:10:17

How do our experimental results confirm or refute this hypothesis?

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RESPONSE -->

confidence assessment: 0

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gk㞲ѷF

assignment #008

008. Using the Acceleration of Gravity

Physics I

06-25-2007

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10:11:31

`q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth.

If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?

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RESPONSE -->

v0=0

'ds=1m

aAve='dv/'dt

we need to know how long it takes to fall that one meter..

v=v0+at

v=0 + at

it seems we lack a little info, either a or t. i'm probably wrong...

confidence assessment: 0

Be sure you respond to the given solution; otherwise it won't appear here in your posted response.

The key is that you do know a, which is equal to the acceleration of gravity.

You would then use the 4th equation of motion to get v0, then either direct reasoning or the 1st or second equation of motion to get `dt.

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10:18:32

`q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump?

Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?

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RESPONSE -->

v0=3m/s

vf=0

a= -9.80/m/s^2

i would use the equation 'ds=vAve * 'dt

so we know the vAve would be 1.5 m/s

all we need to do is figure out the time. and i don't know how to do that

confidence assessment: 0

from v0 and vf you can certainly get vAve, but as you see that doesn't help.

However you can also get `dv. Since a = `dv / `dt and you know a, you can get `dt, and use this with vAve to get `ds.

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10:28:21

`q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table.

If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?

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RESPONSE -->

v0=3m/s

a=9.80m/s^2

'ds=.9m

vf=0

'dt=-3m/s / (9.8m/s^2) = -.3sec

this is what i've figured out so far. now i'm lost a little

confidence assessment: 0

You've just about got it, but not for the right problem. The problem you've almost solved has initial vertical velocity 3 m/s, acceleration of gravity and final vertical velocity 0.

Your v0 = 0 assumes that the upward direction is positive, which is fine. On the other hand, the acceleration of gravity is downward so a = 9.8 m/s^2 assumes that the downward direction is positive. Gotta be one or the other, but it can't be both in the same problem.

Go with the first assumption. Then a = -9.8 m/s^2 and you get `dt = +.3 sec.

Be sure you understand this.

Now stop there because the problem you have framed does not correspond to the given problem.

For the given problem the 3 m/s was horizontal. The a and `ds were vertical. You don't mix vertical and horizontal--the two are independent.

In the vertical direction you've got v0 = 0, a = -9.8 m/s^2 and ds = .9 m. Use the fourth equation of motion to get vf then reason out v0, etc.

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You've made good attempts, but haven't responded to the given solutions, so they aren't included here. There are several disadvantages to this, three of which are listed below:

1. You don't have the given solutions for reference.

2. You haven't gone through the self-critique process, which is the most beneficial thing you can do for yourself.

3. A lot of what I typed in is a repeat of what would be there in the given solution; I enjoyed doing that here, but in order to respond to 100 students I need to confine myself to comments that don't repeat what has already been provided, and I can't do much more of that. For the most part I need to confine my responses to your self-critiques.

See my notes and let me know what questions you have. It might be a good idea for you to run through this again and include self-critiques, etc., so you'll have everything you need on your access page.