Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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. Diagram illustrating problem

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.

share|improve this question
    
Related type of homework problem: physics.stackexchange.com/q/18216/2451 –  Qmechanic May 9 '12 at 19:15
add comment

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
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
add comment

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 $W_\text{fr} = \vec{F}_\text{fr} \cdot \vec{s} = - F_\text{fr} L$.

share|improve this answer
add comment

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!

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.