AP Physics C Mechanics: Conservation of Energy Problem

[SHawking]

Homework Statement

A block of mass m hangs on the end of a cord and is connected to a block of mass M by a pulley arrangement. (m hangs freely, M is attached to the string, and is on a table top. (So m is attached to a string which goes therough a pulley changing the strings direction 90 degrees where it goes through a pulley attached to M and back to a block on the table. So, for every unit m moves M moves two, sorry this is hard to explain without a diagram.) Using energy considerations,
a.) find an expression for the speed m as a function of the distance fallen,
b.) repeat the previous assuming that a friction of µ acts on the block on the table. Assume everything is at rest.

I don't think it is pertinant, but, it has a few latter parts which are calculus based that I was able to get using what I know to be the answer to a and b.

Homework Equations

U=mgh
K=.5mv^2
F=µN (N=mg)
MAybe some kinematics equations for part B?)

The Attempt at a Solution

I think I got part a, just a math error.

Using the formula K(m)+K(M)+U2(m)=U(m) U2 being the new potential energy after a change in distance, and then
1/2mv^2+1/2m(v/2)^2(It will be half the speed of m)+mgh2=mgh
and solving for M, I think I am getting the correct answer

For B, that is including coefficient of friction I think I am having a lot more trouble. My first attempt was to attempt to factor in the friction into the velocity, but that got an incorrect answer. I then tried tog et force, and integrate (integral of NetForce dx=∆K)

Can anyone point me in the right direction?

 

[Delphi51]

I don't see how you can find the acceleration using the energy approach.
I think you will have to work with forces. The force of gravity on m is what makes the masses accelerate. Can you write an F = ma equation for each of the masses? Would the force pulling M be half the force with which m pulls on the rope?

 

[tomcue]

As a scientist, my response to this problem would be to approach it using the principles of conservation of energy. This means that the total energy of the system (including both blocks and the pulley) remains constant throughout the motion.

In part a), we can use the equation for conservation of energy to find the speed of block m as a function of the distance fallen. This can be done by equating the initial potential energy of block m (mgh) to the final kinetic energy of block m (1/2mv^2) plus the final potential energy of block M (mgh2). This results in the equation 1/2mv^2 + mgh2 = mgh, which can be solved for v to get the speed of block m.

In part b), we need to take into account the friction acting on block M as it slides on the table. This frictional force will do work on block M, reducing its kinetic energy and increasing its potential energy. This means that the final potential energy of block M will be less than the initial potential energy of block m (due to the work done by friction), and this needs to be accounted for in the equation for conservation of energy. We can use the equation W = ∆K, where W is the work done by friction and ∆K is the change in kinetic energy of block M. This can be integrated over the distance fallen to find the final speed of block m as a function of the distance fallen, taking into account the work done by friction.

It is also important to note that in both parts a) and b), the system will reach a point where the blocks will stop moving and all the energy will be converted to potential energy. This is when the final kinetic energy of block m (or block M in part b) is zero. At this point, the final potential energy will be equal to the initial potential energy, and we can solve for the maximum distance fallen before the blocks come to a stop.

Overall, this problem can be solved by applying the principles of conservation of energy and taking into account the work done by friction in part b). It may also be helpful to draw a free body diagram and use Newton's laws of motion to determine the forces acting on the blocks.