Friction force incline two masses connected by a string

[konto77]

Homework Statement

It was on my test yesterday. There is a 10kg block on an incline of 45degrees connected by a string to another block hanging over the side of the incline by a frictionless massless pulley. Kinetic friction on the block on the incline is mu=0.5 The block that is hanging over the edge is 10kg. Assume g=10 m/s^2

(a) Find the acceleration of the blocks.
(b) Find the tension in the string
(c) If the hanging block descends by 1 m, what is the work done by friction?

Homework Equations

F=ma Fgxdirection= sin45*mg forceoffriction= .5 * mgcos45

SumFhangingblock= T-mg=ma
SumFinclineblock=T-sin45*mg-.5*cos45*mg

The Attempt at a Solution

I used the two sumF equations i wrote up there ^^ to find "a", but I got -6.07m/s^2. But negative looks wrong to me.

Then I plugged that into one of the sumF equations to get T. I think i got 44.5 N

Then for the work done by friction I multiplied the force of friction by 1m since the block should slide one meter when the hanging block descends one meter.
I'm not sure if I'm right, if not can someone help me out?

 

[Doc Al]

SumFhangingblock= T-mg=ma
SumFinclineblock=T-sin45*mg-.5*cos45*mg

Your second equation is incomplete.

Careful with signs. (Be sure to use a consistent sign convention for the acceleration, depending on whether the block is going up or down.)

 

[konto77]

Oh, is it that I forgot to put SumFinclineblock=T-sin45*mg-.5*cos45*mg this equal to ma?

I did that on the test, skipped my mind here

 

[Doc Al]

Oh, is it that I forgot to put SumFinclineblock=T-sin45*mg-.5*cos45*mg this equal to ma?

OK, fine. Now realize that your two equations have inconsistent signs for the acceleration. Let's assume that the hanging mass falls and thus the sliding mass goes up the incline. In your second equation you use +a for the acceleration, making it in the same direction as the tension. Good! But in your first equation (for the hanging mass) you also use +a for acceleration--but it goes down, opposite to the pull of the tension. Fix that first equation and try again.