What is the speed of block one when block 2 has fallen h?

[Vitani11]

Homework Statement

A block with a mass m1 is held in place on a table and is attached to a block with mass m2 by a massless inextensible string that passes over a massless pulley with frictionless bearings. The coefficient of static friction between block 1 and the table is μ_s and kinetic is μ_k. When block 1 is released, what must be true for block 2 to fall? If that requirement is met, what is the speed of block 1 when block 2 has fallen a distance h?

Homework Equations

Conservation of energy
K_o +U_o -W = K_f + U_f

The Attempt at a Solution

In order for block 2 to fall its force due to tension and gravity has to be greater than the difference between the force of block one due to tension and its force of static friction. This is what I think. Thoughts? Also, how would I put this in terms of conservation of energy?

For the speed I used (1/2)(m₁ +m₂)V² -μ_km₁gh = (1/2)(m₁+m₂)gh and solved for V. I used h for the part of work because the same distance block 2 falls is the distance block 1 moves.

 

[Vitani11]

Here is the set up. Because I defined the potential energy being 0 as the initial height of mass 2 I think there should be a potential on the left side of the equation, but then it would be negative.

 

[haruspex]

In order for block 2 to fall its force due to tension and gravity has to be greater than the difference between the force of block one due to tension and its force of static friction.

That is not right. You might have confused yourself in the process of writing it out in words. Write it as an inequality in algebra.

(1/2)(m1 +m2)V2 -μkm1gh = (1/2)(m1+m2)gh

No. According to that, the greater the friction, the greater the velocity.

 

[Vitani11]

Okay so if I changed the signs in the equation is it then valid? Or am I missing terms still?
And the total energy of block 2 has to be greater than the addition of kinetic energy and work rather than the difference between its force and friction. How is that?

 

[haruspex]

if I changed the signs in the equation is it then valid?

No. Explain the potential energy term.

the total energy of block 2 has to be greater than the addition of kinetic energy and work rather than the difference between its force and friction.

For part 1? No. Take what you wrote initially and turn it into an algebraic inequality.

 

[Vitani11]

Wow holy cow I'm so sorry for that... I've got the right equation - thank you again for your help. I've written the inequality and saw what was not right.