# Principle of Work and Energy

1. Feb 15, 2012

### Peter P.

1. The problem statement, all variables and given/known data
Determine the velocity of the 60- block A if the two blocks are released from rest and the 40- block B moves 0.6 up the in cline. The coefficient of kinetic friction between both blocks and the inclined planes is = 0.05.

Here is the given diagram:

3. The attempt at a solution
Since there is a pulley system, I began with the length equation so that i can determine the displacement of A with respect to B.

ΔsA = -0.5ΔsB
Once everything is plugged in, ΔsA = -0.3 (or just 0.3m down along its plane).
Using this, i calculated the change in height as it falls from the top to the next position. Δh = 0.2598076211 m. Then i calculated the normal force exerted by the plane and the force of friction acting on the block A (Ff = 14.715 N (directed towards the pulley system). So here's the part that i dont get. Im not sure how to tie everything together. I know that:
Kinetic Energy (T), Work and potential Energy (U)
T1 + U1 -> 2 = T2
0.5mAv12 + U1 -> 2 = 0.5mAv22

From the free body diagram of the isolated mass A, i know that there is a normal force, the weight of the block, and a friction force acting on it. So from this information, I believe that there should be gravitation potential energy (VG = mgΔh), and i feel like the friction force needs to be included as well. So i would integrate the force i calculated, with respect to s (∫Ff ds). Then i feel like it should be integrated from 0 -> 0.3. Is this correct so far?

Thanks in advance for any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 15, 2012

### PeterO

Firstly I have never seen Kinetic Energy given the symbol T ? That is usually reserved for Tension.

Secondly, when you looked at just body A, I didn't see any mention of the Tension force in the string??

EDIT: Thirdly, I wouldn't be integrating anything to solve this problem.

3. Feb 15, 2012

### Peter P.

Using T for the kinetic energy is the convention used by my professor and i agree that it was odd when i first saw it as well. Second, i completely forgot about the tension, which only confuses me ever so more in how i need to incorporate it into the equation. Im thinking it depends on mB.

4. Feb 15, 2012

### PeterO

Since you thread title is "Princilpe of Work and Energy" - I would be limiting my throghts to those "global" ideas rather than the nitty gritty of each mass.

You have already deduced the 2:1 ratio between movements of each mass.

I would work on:

Mass A goes down - loses Grav. Potential Energy
Mass B goes up - gains Grav. Potential Energy
Mass A slides its distance against its friction [ μN where N stands for Normal Reaction Force] losing some energy.
Mass B slides its distance against its friction, losing some more energy.

The potential energy gains&losses present a start for the final Kinetic Energy of the masses - but some of that is lost to friction on each block.
What is left should enable you to find v using (1/2)mv2, making sure you use the entire mass, as both will be moving since they are tied together.

Last edited: Feb 15, 2012