A block weighing 50kg is sliding down a slope of gradient 15 degrees. It’s initial velocity, at the point A is 4m/s. Calculate it’s final velocity when it reaches the bottom of the slope, at the point B, 10m away from A. The coefficient of friction between the block and the slope is 0.3. Any help on how to answer this question would be greatly appreciated. Thanks :)

To answer this question, as for all these incline-plane problems, the key is to switch up your x-y plane. Specifically, you want the "hypotenuse'' of the ramp to be the x-axis, and typically it is a good idea to set x=0 at the very top of the x-axis. This determines where your y-axis is.

The gradient of the incline is 15 degrees, and we can see from geometry that the angle between the force vector of the block's weight is 15 degrees away from the line that is formed when you draw the Normal Force of the block on the ramp through the whole body of the ramp .

Now, since the block travels from A to B, it stays exclusively on our x-axis. So, we can neglect y-dimensional motion and forces. We know from general kinematics equations:

Vf^2=Vi^2+2a(xf-xi) ; Vf - final velocity, Vi - initial velocity, a - acceleration, xf - final position, xi - initial postion.

So, we are looking for Vf. We know Vi, and we know xf-xi .

Now, it is just up to finding a. To do this, recognize that we can form a triangle from the weight vector and the extension of the normal force vector and a line between them. This new line between them, which is calculated as mg*sin(15) describes the force of the gravity that applies in the x dimension, now that we are on an incline and have a non-conventional x-dimension. Also, we know that the force of friction, Fr, is calculated as:

Fr=Fn*u ; Fn - normal force , u - coeff of friction

We find Fn from mg*cos(15), again from our triangle.

So, we calculate the overall acceleration from these two forces and we then plug that into our above equation.