Work-energy theorem in relation to the velocity of two masses

[lfused]

Two masses are connecte by a ligh string over a light, frictionless pulley the table surface that m2 is on is also frictionless. Apply the work-energy theorem for this system to calculate the speed of the masses after the masses have moved a distance delta x starting from rest. Note that the work of the tensions drop out. Use the result to obtain the acceleration of the system.

I've looked at this problem for a while now and I just don't know how to get the work of the tensions to drop out given that if m₁ is that mass that's hanging and the forces working on it is F= T₁-m₁g and the forces working on the mass on the table (m₂) is N-mg=ma=0 and T₂-m₂g=F...
yeah. I understand that F=ma and that W=dF= 1/2mv²
but I just can't seem to understand how the tensions fall out of the equasion..

 

[alphysicist]

Hi lfused,

What is the work done by the tension on each of the blocks? How are $T_1$ and $T_2$ related?

Also, is m2 on a table? If so, your horizontal equation for m2 is not quite right.

 

[lfused]

ok here's an image to show what's going on here http://www.geocities.com/liquid_fuse/masses.gif

The mass is a frictionless surface so what other forces could be acting on it on the x-axis?

 

[Doc Al]

Answer alphysicist's questions:

• How do the tension forces on each mass relate to one another?

• Regarding mass 2, what horizontal forces act on it?

 

[lfused]

T1 is the vertical force pulling m1 up and T2 is the horizontal force pulling m2 towards the pulley. And it seems that T2 is the only force that acts on m2..

 

[Doc Al]

And it seems that T2 is the only force that acts on m2..

Yes, T2 is the only horizontal force on m2. But how does T1 relate to T2? (Is one bigger than the other? Which one?)

 

[lfused]

I've already presented all the information given for that problem. Although.. sorry about the typos..

 

[Doc Al]

What you are missing is just a basic fact about the tension in a massless string that loops around a massless and frictionless pulley: The tension is the same throughout.

 

[lfused]

so.. I can just assume that? wow.. that seems too simple..

 

[lfused]

thanks for your help. I really appreciate it. ^_^