# Finding Distance an Object Travels Up an Incline After Launch

I've been doing a review for an introductory physics course final. I have a question on one problem though. Here is the problem:

A mass (M=2kg) is placed in front of a spring with k=900N/m, compressed d=50cm. The spring is released, shooting the mass forward from A to the spring’s equilibrium position at B (A to B is frictionless). The mass then travels along a flat surface from B to C (L=20m), with μk=0.15. At C the surface becomes frictionless and smoothly inclines upwards. The speed of the mass at point D is 0m/s.

What I've been asked so far: velocity at point B, then at point C. I calculated these out and got $10.61$ ${m}\over{s}$ and $7.33$ ${m}\over{s}$, respectively. However, now I'm being asked the height, $h$, of point $D$.

EDIT: Solution
How could I have forgotten about gravatational potential energy!? I used $K_i$ = $U_f$, to retrieve the answer, knowing that $K_i$ = ${1}\over{2}$$mv^2 and U_f = mgh. This also made me notice (as I've been taught before) that mass is irrelevant in this part of the problem. Attempt at solution: I know 3 equations that I assumed would help: x = {1}\over{2}$$at^2$ + $v_{ox}t$ + $x_o$
$v = at$ + $v_{ox}$
$v^2 =$ $v_{ox}^2$ + $2a(x - x_o)$
I know that $v_o$ in any of these would be equivalent to the velocity I found at $C$, and that $v$ would be 0, since the object is at rest at point $D$. However, I don't know any times nor accelerations (as far as I'm concerned), so I'm stuck.
Other info: we've learned about conservation of energy and momentum, work, power, and a few others, but simple kinematics seem to be the only appropriate application here, unless I'm wrong on that too.

The answer is supposedly $2.74m$, but once again, I'm unsure how to get here. Any pointers would help, as I'd like to be well prepared for tomorrow's final!

Relevance of question (in my opinion): yes, this is a specific question, which isn't necessarily favored, but this problem seems rather common in introductory physics courses, so I'm sure many others that need help could see this and apply it appropriately as well.

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Related type of homework problem: physics.stackexchange.com/q/18216/2451 –  Qmechanic May 9 '12 at 19:15

we've learned about conservation of energy

Yes.. this is what you should look at. Since the surface is frictionless from C to D, you will not lose any energy due to friction. So the sum of kinetic and potential energies will be constant, and equal to the value of the total energy at C. This should be enough to work out the height. Remember the expression for the potential energy of a particle of mass $m$ placed at a height $h$ when the acceleration due to gravity is $g$, and you should get the answer.

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How could I forget about gravatational potential energy! Alas, I understand now :). I had to use spring potential energy to find velocity at point B, so I forgot about the other forms of potential energy.. Final exam stress is killing me! –  Mike Gates May 9 '12 at 19:22
@MikeGates The easiest way is to forget kinematics of the problem completely and do it all with energy (of course not forgeting work done by non-conservative forces). Just taking initial and final energy. Try it, it's twice as easy. –  Pygmalion May 9 '12 at 19:24

The easiest way is using modified law of conservation of energy that states

$$W_\text{NC} = E_\text{final} - E_\text{init}$$

where $W_\text{NC}$ is work done by non-conservative force(s) (in this case friction).

All you need is to calculate initial energy (potential energy of spring), final energy (gravitational potential energy) and work done by friction $W_\text{fr} = \vec{F}_\text{fr} \cdot \vec{s} = - F_\text{fr} L$.

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The potential energy of the spring is $1/2 Kx^2 = 0.5\times 900\times 0.5^2 = 112.5\mathrm{J}$

The energy lost in the section with friction (B to C) is the work done on the object by friction:

$$W = (Fd) = (u N d) = (u m g d) = 0.15\times 9.81\times 2\times 20 = 59\mathrm{J}$$

The energy available to raise the block is then $112.5 - 59 = 53.5\mathrm{J}$

$$PE = mgh$$ so $$h = PE/(mg); h = 53.5/(2\times9.81) = 2.73 meters$$

N ice problem!

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