Denote the angle of inclination as alpha.
Let's draw a free-body diagram. The forces acting on our block are:
1) the gravity force, `mg,` acts down;
2) the traction force, `F_T,` acts upwards along the inclined plane;
3) the reaction force, `N,` act upwards perpendicular to the inclined plane;
4) the friction force, `F_f,` acts downwards along the inclined plane.
We know `F_f=mu*N,` where mu is a coefficient of friction.
Together they make the net force `F,` which is equal to `ma,` where `a` is the acceleration of the block (by Newton's Second Law). Because the block moves along the inclined plane, `a` is also along this plane.
Please look at the picture. We'll use x-axis parallel to the inclined plane and y-axis perpendicular to it.
Consider projection on the y-axis:
And on the x-axis:
Substitute `F_f=mu*N=mu*m*g*cos(alpha)` into the second equation:
So `a=F_T/m - g*(mu*cos(alpha)+sin(alpha))`
which is `approx 32/4-9.8*(0.115*0.82+0.57) approx 8-6.5=1.5 (m/s^2).`
This is the answer to the second question, what is the acceleration of the block up the incline.
The friction force is `mu*m*g*cos(alpha) approx 3.7(N).`
This is the answer to the first question.
If the rope breaks, the equation will be (`F_T=0` , `N` and `F_f` remain the same)
`a= - g*(mu*cos(alpha)+sin(alpha)) approx -6.5(m/s^2),`
or 6.5m/s^2 down the incline. This is the answer to the third question.