Thanks for looking at this post.
Here was my though process as I approached this problem...
We'll just use the conservation of energy theorm for this problem.
Einitial=Efinal
Kinitial+Uinitial=Kfinal+Ufinal
Since the massive object has no initial kinetic energy at the start of the system, call it when the object is at its initial height, it has potential energy equal to mgh. m is the mass of the object, g is the acceleration due to gravity, and h is equal to the distance the object travels. In this case, I have to set h equal to a change in the initial height if the object to its height when it encounters the frictional surface. Δy=h. The left side of the equation now reads: Uinitial=mgh. Or just, mgh
mgh=some final kinetic energy - the work done by friction
We have an equation that relates kinetic energy with velocity...it is 0.5mv2=Kfinal...v2 is also the final velocity, which for this system, is right at the point when object leaves the surface. The object will travel in the air for some unknown distance at some other velocity, but for purposes of this problem, we won't worry about them, because those things are not what this problem is asking for. Now, I'm not sure about multiplying, by length d, but I believe we have to.
I also know that the work done by the frictional force equals μNd. That's because work equals force x distance. We have μ and d given to us. d in this case is the distance the object travels along the rough surface only. N is unknown, but we do know that in this case N =W, which happens to = mg in this case. I may not have told you that the rough surface is flat. The minus sign comes in because the frictional force opposes the motion of the object.
Now our equation looks like this
mgh = 0.5mv2d - 1.532622
Now let's isolate v. My m will cancel when I move everything over to the other side.
(gh + 1.532622)/(0.5)(d) =v2
PlUg it all in and take the square root...v = 5.96 m/s
And sorry that the + is underlined..I'm not sure why it does that when I type it in.