Calculating Work by Tension on Masses

[Yae Miteo]

Homework Statement

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?

Homework Equations

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

$$w=Fx$$

The Attempt at a Solution

[/B]
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.

 

[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?

 

[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?

 

[Doc Al]

so
$$T=m_2g$$

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

 

[Yae Miteo]

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

 

[Doc Al]

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

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.)

 

[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?

 

[Satvik Pandey]

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

 

[Doc Al]

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?

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

 

[Satvik Pandey]

So, solving for tension, I get

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

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

and then putting it into$$W=Fd$$

I get

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

Is this correct?

I think this is correct.

 

[Doc Al]

So, solving for tension, I get

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

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

 

[Yae Miteo]

yes, it was a typo

 

[Yae Miteo]

And the 2nd law equations for tension are

$$F_1=m_1a$$

$$F_2=m_2g$$

right?

 

[Satvik Pandey]

What is F₁ and F₂?

 

[Doc Al]

And the 2nd law equations for tension are

$$F_1=m_1a$$

$$F_2=m_2g$$

right?

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.

 

[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
$$T = \cfrac{m_1m_2g}{m_1+m_2}$$

 

[Satvik Pandey]

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

That looks right.

 

[Yae Miteo]

So then into
$$W=Fd$$

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

 

[Satvik Pandey]

So then into
$$W=Fd$$

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

This is correct.:)

 

[Doc Al]

I worked it like this:

tension will equal the sum of both forces, so
$$m_2a + m_2g = T$$
and
$$T = m_1a$$

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.

 

[Yae Miteo]

Thank-you so much guys! I got it figured out.

 

[Satvik Pandey]

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.

Yes, there should be negative sign.

 

[Satvik Pandey]

Thank-you so much guys! I got it figured out.

You are welcome.:)