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)...

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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:

`N=m*g*cos(alpha).`

And on the x-axis:

`F_T-F_f-m*g*sin(alpha)=ma.`

Substitute `F_f=mu*N=mu*m*g*cos(alpha)` into the second equation:

`F_T-mu*m*g*cos(alpha)-m*g*sin(alpha)=ma.`

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.**