Physics Help

[minime72706]

Here's the problem-

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless peg to a 2.2 kg mass m2 that hangs over the side of the shelf 1.5 m above the ground (Figure 5-62). The system is released from rest at t = 0 and the 2.2 kg mass strikes the ground at t = 0.86 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 2.2 kg mass now strikes the ground 1.3 seconds later.
Pic of problem

I need to find m1 and the coefficient of kinetic friction between m1 and the shelf.

I've worked at this for a while, and have yet to find the correct answer. There isn't a similar problem that I can find to work from. Its for a practice midterm, and I'm assuming it or something similar will be on the test, so I would like to know how to do it. Any help is greatly appreciated.
Thanks

 

[FleshLight]

Looks like you need to use a kinematic equation to find the acceleration (since you have a dY and a dT). Then you draw a free body diagram for M1:

Normal Force
^
|
|
M1 <------Friction, ------> Tension
|
|
v
mG

You can find the acceleration of m2, you know its mass, then you can determine the tension. Then just plug everything into "sigma F = 0" and you win!

 

[darthsidious]

Write down two equations. An assumption I'm making here is that the string has tension, i.e it does not slack,which should be valid.

Assuming tension on string is T. then:
m2*g - T = m2*a, where a is the downward acceleration of the block.
T - mu*m1*g = m1*a, where a is the rightward acceleration of m1 (same magnitude as m2 ifstring is taut.
s(distance travelled) = 0.5 a*t^2 (as inital velocity is zero). Solve for a here, and solve for m1 from the above two equations.

Hope this helps

 

[RGUN]

Determine the acceleration of the second part with the 1.2kg mass involved. With that you can find the force (f=ma). From that you can determine the friction force, since the total force will be a product of both friction and weight only. Now that you have frictional force you can determine the coefficient of friction. Apply that to the first part of the problem and come up with a mass.

 

[minime72706]

So am I correct in assuming the add mass is unnecessary to the problem? My math must be going wrong someplace, because I came up with mass=3.05 and mu=.422, which the practice test said was wrong.
I did the following:
.5at^2=d
so a=4.056m/s^2
T=12.65
mu=T/m1-a/g
I plugged that into my Ti89 which gave me the expression:
40.55923*m1-11.416=4.056m1
Then I solved for m1 and mu.
But something was wrong.

 

[darthsidious]

Originally posted by: minime72706
So am I correct in assuming the add mass is unnecessary to the problem? My math must be going wrong someplace, because I came up with mass=3.05 and mu=.422, which the practice test said was wrong.
I did the following:
.5at^2=d
so a=4.056m/s^2
T=12.65
mu=T/m1-a/g
I plugged that into my Ti89 which gave me the expression:
40.55923*m1-11.416=4.056m1
Then I solved for m1 and mu.
But something was wrong.

How did you come up with that number for tension? did you plug a into the other equation?

 

[minime72706]

Originally posted by: darthsidious

Originally posted by: minime72706
So am I correct in assuming the add mass is unnecessary to the problem? My math must be going wrong someplace, because I came up with mass=3.05 and mu=.422, which the practice test said was wrong.
I did the following:
.5at^2=d
so a=4.056m/s^2
T=12.65
mu=T/m1-a/g
I plugged that into my Ti89 which gave me the expression:
40.55923*m1-11.416=4.056m1
Then I solved for m1 and mu.
But something was wrong.

How did you come up with that number for tension? did you plug a into the other equation?

I used m2*g - T = m2*a. I got 12.63. the .65 was a typo.