# Work by tension on masses

1. Sep 26, 2014

### Yae Miteo

1. The problem statement, all variables and given/known data

A mass m_1 = 4.1 kg rests on a frictionless table and connected by a massless pulley to another mass m_2 = 3 kg, which hangs freely from the string. When released, the hanging mass falls a distance d=0.83m.

How much work is done by tension on m_1?

2. Relevant equations

$$v^2 = v_0^2 + 2a(x-x_0)$$

$$w=Fx$$

3. The attempt at a solution

I began with w=Fx. I re-wrote it as

$$w=max$$

because F=ma.

But from here I'm stuck. I can't find acceleration, and I'm pretty sure it has something to do with gravity, but not completely. I feel that I'm pretty close but I just can't get that last part.

2. Sep 26, 2014

### Satvik Pandey

Hi Yae Mieteo.
Draw the free body diagrams of m1 and m2 and set up equations using 2nd law.
Can you find tension in the string?

3. Sep 27, 2014

### Yae Miteo

After doing that, I get this.

$$F=ma$$
so
$$T=m_2g$$
to get a, I solve F=ma to get a=F/m (and m is m_1 + m_2), and then plugin into
$$W = Fd$$
I get
$$W=mad$$
so
$$W= \frac{(m_2)(T)(d)}{m_1 + m_2}$$
and plugging in I get
$$W=10.3177$$
which is wrong. I think I'm even closer, but there is an error somewhere, maybe wrong masses?

4. Sep 27, 2014

### Staff: Mentor

If that were true then the net force on $m_2$ would be zero and it would not accelerate.

5. Sep 27, 2014

### Yae Miteo

Would tension then be
$$T=m_1g$$
or would it involve both mass 1 and mass 2?

6. Sep 27, 2014

### Staff: Mentor

It seems like you're just guessing. Instead, set up two equations--one for each mass--and solve for the tension. (Yes, it will involve both masses.)

7. Sep 27, 2014

### Yae Miteo

So, solving for tension, I get

$$T = \cfrac{m_1m_2g}{m_1-m_2}$$

and then putting it into

$$W=Fd$$

I get

$$W=\cfrac{m_2m_2gd}{m_1+m_2}$$

Is this correct?

8. Sep 27, 2014

### Satvik Pandey

Can you set up equations using 2nd law for M1 and M2?

9. Sep 27, 2014

### Staff: Mentor

Yes, that looks good. (Except for the minus sign in your expression for tension. A typo?)

Last edited: Sep 27, 2014
10. Sep 27, 2014

### Satvik Pandey

How you got minus sign in denominator?
I think this is typo.
I think this is correct.

11. Sep 27, 2014

### Staff: Mentor

I missed that minus sign. I assumed it was a typo. Was it?

12. Sep 27, 2014

### Yae Miteo

yes, it was a typo

13. Sep 27, 2014

### Yae Miteo

And the 2nd law equations for tension are

$$F_1=m_1a$$

$$F_2=m_2g$$

right?

14. Sep 27, 2014

### Satvik Pandey

What is F1 and F2?

15. Sep 27, 2014

### Staff: Mentor

No. All second law equations should be in the form of $\Sigma F = ma$.

How did you solve for the tension before? Not this way.

16. Sep 27, 2014

### Yae Miteo

I worked it like this:

tension will equal the sum of both forces, so
$$m_2a + m_2g = T$$
and
$$T = m_1a$$
solve for a
$$a = \cfrac{m_2g}{m_1 + m_2}$$
plug in
$$T = \cfrac{m_1m_2g}{m_1+m_2}$$

17. Sep 27, 2014

### Satvik Pandey

That looks right.

18. Sep 27, 2014

### Yae Miteo

So then into
$$W=Fd$$

$$W=\cfrac{m_1m_2gd}{m_1 + m_2}$$

19. Sep 27, 2014

### Satvik Pandey

This is correct.:)

20. Sep 27, 2014

### Staff: Mentor

You have a sign error. Note that if you actually solved these equations as written, you'd get a minus sign instead of a plus sign in the denominator of your expressions for a and T.

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