Melting ice block on an inclined plane?

[NotANumber]

Hello all...

Help...

my teacher gave us this question and...

no matter how I approach this question, I just can't solve it...

The question asks...

an ice block melts and slides down a plane of rough surface..

The mass of the ice block starts at 3 kg, and its mass decreases as it melts..

The ice block decelerates at -0.2 m/s^2 per 0.2 kg mass starting at an acceleration of 0.7 m/s^2... (the incline of the plane is 20 degrees, so the angle between the Fgravity and Fperpendicular is 20 degrees as well, Fperpendicular = Fpull, that I know)

and the question asks me to find the mass of the melting ice block at 0m/s^2 and as well the friction acting on the ice block (assumed to be constant)

I think I kinda know how to do this (I can do mass on an inclined plane with u coefficient), but my teacher doesn't like giving us u... plus I just kinda can't think my way around this melting mass with an shifting acceleration thing...

Help?

 

[Stonebridge]

I think the wording of this question could be a little clearer. Particularly, exactly what is meant by "decelerates at -0.2m/s/s per 0.2kg"
I take it to mean that the block, initial mass 3kg, starts slipping down the plane and has an initial acceleration of 0.7m/s/s down the plane. It is melting, and for every 0.2kg of mass it loses, the acceleration reduces by 0.2m/s/s.
So when it is 2.8kg the acceleration is 0.5m/s/s, when it is 2.6kg the acceleration is 0.3m/s/s and so on.
So what is the mass when the acceleration is 0m/s/s

When the acceleration is zero, what do you know about the resultant force on the block?
If you know the component of the weight down the slope, you can find the frictional force required to give zero acceleration.
I imagine the reason your teacher does not want to discuss the coefficient of friction, is that in order for this all to happen as described, its value cannot remain constant.