Modelling the height of a block up a slope with time

[oMattz]

1. Homework Statement

A block, of mass M1, on the rough slope shown is attached to another mass M2 by a light, inextensible string which passes over a frictionless pulley as shown in the diagram (the coefficient of friction between the block and the slope is u ). The block is released from rest. Find an expression for the height of the block above the bottom of the slope as a function of time. (You should consider the full range of behaviour that may happen for different values of the parameters).

Homework Equations

[/B]
F(gravity down slope) = mgsin(theta)
F(friction down slope) = umgcos(theta)
f=ma
s = ut +1/2at^2

The Attempt at a Solution

Force of gravity on block is m1gsin(theta)
Force of friction on block is um1gcos(theta)
Force needed to push block up slope at constant speed = m1gsin(theta) + um1gcos(theta)
If M2g > Force needed then block will accelerate up slope with a=(M2g-(m1gsin(theta) + um1gcos(theta)))/m1
vertical component of that acceleration = asin(theta)
height = ut + 1/2at^2
height = 0.5*(m2g-(m1gsing(theta) + um1gcos(theta)))/m1)sin(theta)*t^2
I'm pretty sure this answer is wrong as it is so complex, but would just like either confirmation that it is right/wrong and where to go

 

[haruspex]

Force of friction on block is um1gcos(theta)

The problem statement does not discriminate static and kinetic coefficients, but it does ask you to consider all possibilities. What would be a more circumspect statement than the above?

If M2g > Force needed then block will accelerate up slope

True. What if it isn't, though?

with a=(M2g-(m1gsin(theta) + um1gcos(theta)))/m1

Don't forget M2 will accelerate too. Consider the tension and analyse the forces on each mass separately.