Finding Speed of a Mass in a Frictionless System at a Specific Time

[Becca93]

Homework Statement

The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are:
M1 = 2.00 kg
M2 = 7.00 kg
M3 = 4.00 kg
Calculate the speed of M2 at a time 1.750s after the system is released from rest.

(Question and diagram attached)


The attempt at a solution

I followed the way my professor did a similar question in my notes, however the answer is not correct. My process was as follows:

For mass 1:
ƩFnet = M1a
T1 - M1g = M1a
T1 = M1(a+g)
T1 = 4(a+9.8)

For mass 2:
ƩFnet = M2a
T2-T1 = M2a
T2 = M2a - T1
T2 = 7a + 2a - 19.6
T2= 9a - 19.6

For mass 3:
ƩFnet = -M3a
T2 - W3 = -M3a
T2 = W3 - M3a
T2 = 39.2 - 4a

I then made the two T2 equations equal to find acceleration of the system:
9a - 19.6 = 39.2 - 4a
9a + 4a = 58.8
a = 58.8/13
a = 4.523 m/s^2(right)

Then I found velocity of M3

Vf = Vit + (1/2)at^2
Vf = (0)(1.750) + (1/2)(4.523)(1.750^2)
Vf = 6.926 m/s


Anyone know where I went wrong? I would seriously appreciate any help. I've tried this a bunch of times and have gotten different answers for acceleration every time, all of them wrong. I only have a few more tries left. Any guidance would bed a great help!

 

[Doc Al]

Then I found velocity of M3

Vf = Vit + (1/2)at^2

You mixed up your formulas. That's the one for distance. You want:
Vf = Vi + at

 

[Becca93]

You mixed up your formulas. That's the one for distance. You want:
Vf = Vi + at

I'm still not getting the correct answer. I must have messed something up before that too. I just can't see where I messed up.

 

[Doc Al]

For mass 1:
ƩFnet = M1a
T1 - M1g = M1a
T1 = M1(a+g)
T1 = 4(a+9.8)

M1 = 2 kg, not 4.

I suggest you not plug in any numbers until the very last step. Solve it with symbols first, then plug in the masses to get the final answer.

 

[Becca93]

Okay, so I've done this question a bunch of times. Like, a lot of times. I only have 2 tries left to get the correct answer.

1 Incorrect. (Try 1) 8.12 m/s
3 Incorrect. (Try 2) -8.12 m/s
4 Incorrect. (Try 3) 2.31 m/s
5 Incorrect. (Try 4) 6.93 m/s
6 Incorrect. (Try 5) 2.28 m/s
7 Incorrect. (Try 6) 14.8 m/s
8 Incorrect. (Try 7) 4.18 m/s
9 Incorrect. (Try 8) 6.87 m/s

I've attached a picture of another failed attempt.

Please, does anyone know how to do this correctly?

 

[Doc Al]

Please, does anyone know how to do this correctly?

Did you correct the mistake I pointed out? Please repost your new solution. (A scan makes it hard to point out any errors.)

FYI: Your three equations in post #1 are correct, so that's a good start.

 

[Becca93]

Did you correct the mistake I pointed out? Please repost your new solution. (A scan makes it hard to point out any errors.)

FYI: Your three equations in post #1 are correct, so that's a good start.

I did, but I wasn't really using the T1 system of numbers because if I let equations one and two equal each other, I ended up with two values I didn't know, and I would have to solve for T2 and equate that new equation to equation 3 to find acceleration.

I thought if I did that, I would definitely end up making a mistake, as I was looking for the velocity of M3.

Also, for clarity, do you want me to type out what was on the scan of the paper?

 

[Doc Al]

I did, but I wasn't really using the T1 system of numbers because if I let equations one and two equal each other, I ended up with two values I didn't know, and I would have to solve for T2 and equate that new equation to equation 3 to find acceleration.

I thought if I did that, I would definitely end up making a mistake, as I was looking for the velocity of M3.

The first thing to do, as you attempted in post #1, is to solve for the acceleration. You have three equations and three unknowns, so it works out OK. You can solve them any way you wish.

Also, for clarity, do you want me to type out what was on the scan of the paper?

Why not just cut and past your work from post #1, making any corrections needed? You were on the right track. (If you want, you can type out what was in the scan.)

 

[Becca93]

The first thing to do, as you attempted in post #1, is to solve for the acceleration. You have three equations and three unknowns, so it works out OK. You can solve them any way you wish.

Why not just cut and past your work from post #1, making any corrections needed? You were on the right track. (If you want, you can type out what was in the scan.)

If I was on the right track with the first one, I'll go from there.

Right now, I'm doing what you said earlier and waiting to put the values in until very last. I've got two tries left before I'm locked out with zero marks, so would you mind telling me if I'm on the right track?

So I had:
T1 = M1(a+g)
T2 = M2a - T1
T2 = M3(g-a)

M2a - M1a + M1g = M3g - M3a
M2a + M3a - M1a = M3g - M1g
a(M2 + M3 - M1) = M3g - M1g
a = (M3g - M1g)/(M2 + M3 - M1)

a = [(4)(9.8) - (2)(9.8)] / (7+4-2)
a = 2.1777777

And then, of course, Vf = Vi + at.

So,
Vf = 0 + (2.177777)(1.750)
Vf = 3.811111

 

[Doc Al]

So I had:
T1 = M1(a+g)

OK.

T2 = M2a - T1

Not OK. Fix this.

T2 = M3(g-a)

OK.

M2a - M1a + M1g = M3g - M3a
M2a + M3a - M1a = M3g - M1g
a(M2 + M3 - M1) = M3g - M1g
a = (M3g - M1g)/(M2 + M3 - M1)

You are very very close. Fix the error above and try again.

 

[Becca93]

You are very very close. Fix the error above and try again.

Oh, oh, okay. So would it be:

T2 = M2a + T1

Making it
M2a + M1a + M1g = M3g - M3a
M2a + M3a + M1a = M3g - M1g
a(M2 + M3 + M1) = M3g - M1g
a = (M3g - M1g)/(M2 + M3 + M1)

a = (19.6)/(13)
a = 1.5079

Vf = (1.5079)(1.750) = 2.638

Does that correct the mistake I was making?

Edit: I submitted it and it was accepted! Thank you so much!

 

[Doc Al]

Perfect.

 

[Doc Al]

Just for fun, here's a 'trick' I use to solve this quickly. If you write the equations this way:

T1 - m1g = m1a
T2 - T1 = m2a
m3g - T2 = m3a

Just add the equations and the tensions cancel.

(Of course you can't count on such a trick working for all problems.)