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**HallsofIvy** I answered (a) then saw that you had already done that! Very good.

For (c), a "range" of values for $\displaystyle \mu$ must have both an upper and lower bound. The upper bound is easy: $\displaystyle \mu$ cannot be so large as to have no motion at all. The total force on the system is $\displaystyle mg(3\mu- 1)$ (you must have calculated that when you did (a)) and the masses will not move if that is negative $\displaystyle mg(3\mu-1)< 0$ leads to $\displaystyle \mu< 1/3$. Now you are told that the mass A, which, as you calculated for (b) has speed $\displaystyle \sqrt{\frac{gh(1-3u)}{2}}$ when B stops falling and so the only force on mass A is the friction force $\displaystyle 3mg\mu$. The acceleration due to that is $\displaystyle 3g\mu$ and that must stop A before it reaches the pulley.

A is initially distance 2h from the pulley. After B has fallen a distance h, A will have moved distance h also and so will be distance h from the pulley. Using decceleration $\displaystyle 3g\mu$ to determine the time required for A to stop. Calculate the distance A will have traveled. Find the value of $\displaystyle \mu$ so that distance would be exactly h. That is the lower bound for $\displaystyle \mu$.