Question

Problem 1: In the figure to the right, m1=20.0kg and α=53.1o . The coefficient of kinetic friction between the block and the incline is µk=0.40. a) What must be the mass m2 of the hanging block if it is to descend 12.0 m in the first 2.00 s after the system is released from rest? b) For the µk in the problem, what mass m2 will provide constant velocity of m1 down the incline? c) Suppose µs is 0.7, what is the maximum mass m2 that will still cause no movement?

Answer 1

Note: The "figure" was not included, however, by looking at the problem i believe i have the right idea of how does it looks. I included a picture for what i assume is the problem, if your problem is not like the picture please let me know since then the calculations should be different than yours since mine are based on this next picture.

P.S: If in your problem the triangle points to the other side the problem will be ok. It will simply change the Free body diagrams and the starting formulas but the end result will be the same.

Ok, now before we start we need to make a free body diagram of both masses

  

Now starting with question a we need to fin the system acceleration, so we do this:

So now we can write down the sum of the forces for the Y axis of M₂, and we have:

This formula (that ill name FORMULA 1) will give us the answer but we dont know T so we calculate it using M₁ so we write down the sum of forces on the X axis.

And on this formula (FORMULA 2) we dont know F_k which can be calculated using the sum of forces on the Y axis. So:

So:

Which means that:

Now returning to the FORMULA 2 we have:

And introducing this value into FORMULA 1 we have:

Answer a: m₂=85.21Kg

Now for question B we need to understand that it will all be the same except that F_k will be going the other side, since m₁ will now be going down, so it will be negative on formula 2. Also Acceleration will be 0 in both formula 1 and formula 2.

So we take FORMULA 2 with the changes i said and we have:

And this new tension on the new FORMULA 1 will give us:

Answer B: If the mass of m₂ is 11.19Kg m₁ will start to move down and for any lower mass m₁ will start to have acceleration

For question c we only have to use but this time the sum of forces will be equal to zero so we calculate F_s and recalculate with the new Formula 1 and Formula 2 which now have F_s which is positive since the m₂ is going down:

This value on new Formula 2 gives us:

And this new tension on the new FORMULA 1 will give us:

Answer C: m₂=65.21Kg