Finding max height of a mass launched by a spring? I have the following problem on my homework. The actual problem has numbers, but I figure using symbolic notation will help me understand the concept in the long run.
A block of mass $M$ is launched by a spring of negligible mass and constant $K$ along a frictionless surface. Prior to launch the spring is compressed a distance of $D_1$. Some time after launch mass then travels up an incline (the angle doesn't matter) that is also frictionless. However, at some point between leaving contact with the spring and entering the incline, the block passes over a rough patch of of distance $D_2$ where the coefficient of Kinetic friction is $C$. Find the max height the mass will travel up the incline. 
I've managed to work through the following:
Looking at this I believe it is a conservation of energy problem. We have the Initial energy is equal to energy final - energy lost through friction.
So, Initial Energy is 
$$E_i = \frac 1 2 kx^2 $$
Then our final Energy is simple $MGH$ since at it's max height it stops and all energy is potential again. 
So $MGH = \frac 1 2 kx^2 - \text{Friction}$
and this is where I get confused. I'm not to sure how I calculate the energy loss of friction.
I believe it is the work friction does is equal to energy loss.
So that would $W=FD$  distance is $D_2$ and the force would be $CMG$ since the normal force is equal to gravity on the flat surface. Am I correct in this?
This would lead me to
$$H = \frac{(1/2kx^2)-(CMGD_2)}{MG}$$
Which simplifies down to:
$$H=\frac{1/2kx^2}{MG} - (CD_2)$$
This also confuses me, if my above calculations are correct, then the amound of energy loss due to friction does not depend on the mass of the object?
 A: Step by step you have:


*

*Potential Energy in compressed spring $T_1=\frac{1}{2} K D_1^2 $

*Kinetic Energy after leaving the spring $K_1 = \frac{1}{2} M v_1^2 = T_1 = \frac{1}{2} K D_1^2$ } $v_1=\ldots$

*Work lost by rough patch $W =(C M g) D_2$

*Final kinetic energy after patch $K_2 = K_1 - W = \frac{1}{2} K D_1^2 - (C M g) D_2 $

*Potential Energy on top of ramp $T_2 = M g H = K_2$


result
$$ \boxed{H = \frac{ \frac{1}{2} K D_1^2}{M g} - C D_2} $$
which is what you have. Good job!
A: First of all from dimensional considerations there is no need for height to have some relation with mass or gravitational force, it must be related to some sort of distance and thats what your equatoon give and that is ok.
Now your question particulary your confusion is of the sort where you have found thay ;
F × y = k + F × x, clearly on simplifying such an equation you would not expect y to be related to F and not just x. But i believe it would be more understandable from the following: 
When two equal forces act for two given distances and the work that has been done comes out to be equal, if no other force existed for the given experiment and thr displacement had taken place in the direction of the force only; we can conclude that for work to be same distances must be same.
There is no need for either of the distances to be anyhow related to the applied force, similarly the force in your question atleast in magnitude is same (also friction is just a constant multiplied by gravitational force) So it is expected and obtained that the height is only related to distance of the rough patch and not the mass of object.
