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Spring, friction, & Energy all in one problem

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    The spring shown in the figure is compressed 50 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the 30 degrees incline is 0.15.

    Part a : Whats the students speed after leaving the spring?
    Part b : How far up the incline does the student go

    knight_Figure_11_54.jpg

    2. Relevant equations

    V = square root ( k * x^2 / m) : 14.1 m/s (answer to part a)

    Conservation of energy

    3. The attempt at a solution

    I was able to calculate part a and came to an answer of 14.1 m/s. For part b, i solved for the equation of : S = (1/2mV^2 + mgY) / mk*mg*cos(theta) = 155 m. That just seems to large of a distance to travel up an incline. Any ideas?
     
  2. jcsd
  3. Nov 29, 2009 #2
    Part a is solved correctly.

    For part b, use conservation of energy again. What are the energies of the students at the top of the hill? Kinetic+potential. The same amount of energy will be present at the bottom of the hill, but now that energy will be all kinetic. Now the student will be going up the hill, therefore you will be taking away energy until the student stopped, or the kinetic energy is 0. Another way of saying that is that all the energy is converted to potential or lost through friction.

    Hope this helps.
     
  4. Nov 29, 2009 #3
    Heres where i'm confused:

    s = the hypothenuse of the hill (what were solving for i guess)

    1/2mv^2 = mgh + W'friction

    h = s*sin(theta)

    W'friction = Mk*FN = Mk*mg

    Plug everything in and solve for s is what i was thinking?
     
  5. Nov 29, 2009 #4
    You have the basic concept correct, just a couple minor flaws in it.

    -the v in your 1/2mv^2 is the speed at the bottom of the hill, which is different from the speed at the top of the hill, make sure you readjust that with potential energy.

    -W'friction=Mk*FN. I'm assuming Mk is the coefficient of friction? Should probably use [tex]\mu[/tex]. But also the FN is not simply mg! Because the student is going up the incline, remember to readjust FN according to the actual force perpendicular to the slope. (Hint: use a trig function)

    Using those should get you the appropriate h, and then s.
     
  6. Nov 29, 2009 #5
    Ok, after reading/messing with your hints, and thinking some more, this is the equation i came up with :

    S (Hypothenuse) = (1/2mv^2 - Mk*mg*cos(theta)) / (mg*sin(theta)) = 20 meters

    Seems logical. Now, the velocity in (1/2mv^2) should still be the 14.1 m/s i found in part a because the speed at the bottom of the hill on a frictionless surface should remain constant, right?

    Thanks Again
     
  7. Nov 29, 2009 #6
    The last part here is incorrect, the speed at the bottom of the hill will not be the same speed at the top of the hill. At the top of the hill, the student possesses kinetic energy from the spring and potential energy from being at the top of the hill. At the bottom of the hill, are there any potential energy left? Where did the potential energy go? :]

    That being said, once you adjust the velocity at the bottom of the hill, you should get the correct answer. The equation you gave seems to be correct.
     
  8. Nov 29, 2009 #7
    I understand your point, but what's confusing me is at the bottom of the hill, our height is unknown (assuming mgh = 0). Therefore, if i use the conservation of energy = 1/2mv^2 = mgh, im going to get 0 which cant be correct for v.

    Energy Initial = Energy Final

    1/2mv^2 + mgh = 1/2mv^2 + 0 ?
     
  9. Nov 30, 2009 #8
    Because you don't know the height at the bottom of the hill, it would be wise to set h=0 at the bottom at the hill, so you would get mgh=0, exactly what you have. I'm not really sure where your confusion is coming from, so I'm just going to take a wild stab here and tell you that the 2 v's are not the same. Meaning the energy conservation equation would look like this:

    1/2m(v1)^2 + mgh = 1/2m(v2)^2

    where v1 is the 14.1 and the v2 is the speed at the bottom of the hill. On the LHS h is 10 and on the RHS there is no mgh since we let h=0 at that point. Notice the mass just cancels out nicely.

    Edit: After thinking about your question again I think I might see the source of your confusion. Potential energy comes from height. So at the bottom of the hill mgh=0, while at the top of the hill mgh=(100)(9.8)(10). I don't really see your logic of setting mgh=0 to 1/2mv^2, mgh(at the bottom of the hill)=/=1/2mv^2 at the bottom of the hill. But rather it's the sum of mgh and 1/2mv^2 at the bottom of the hill= the sum of mgh and 1/2mv^2 at the top the hill. But then you seemed to corrected your own contradiction with the correct equation for energy conservation. So I'm just thinking you're confusing the two different v's.

    But try to think about it in a different light, you're using the conservation of energy here. So is the speed at the bottom of the hill really relevant to the question? Is finding the speed at the bottom of the hill necessary to solve the problem? What you are looking at here is conservation of energy. We know that the sum of all energy (including potential) at the top of the hill is the same at the bottom of the hill (where it is all kinetic). Pretty obvious from the earlier equation:

    1/2m(v1)^2 + mgh = 1/2m(v2)^2

    Now, let's look at the equation you have earlier for S:

    S (Hypothenuse) = (1/2mv^2- Mk*mg*cos(theta)) / (mg*sin(theta))

    From the construction of your equation, you're saying the initial energy is only kinetic, therefore the v here should have been v2. That's the fundamental mistake in your first attempt, you assumed the v here is v1, but if you used v1, you're not including the gravitational potential energy that should have been there. To take this one step further, you also have:

    1/2m(v1)^2 + mgh = 1/2m(v2)^2

    Where h here is the first hill, so 10, m=100kg, and v1 is readily available from your answer to part a. Do you see a nice substitution here that can bypass the need to find v2?

    A lot of things here, sorry if I sound confusing >_>
     
    Last edited: Nov 30, 2009
  10. Nov 30, 2009 #9
    That is probably the best explanation anyone could write on a forum! I was confused on whether to add mgh to the equation or not and obviously i should have went with my gut feeling. Thank you for that explanation. I'm also extremely tried if that has any justification for all my questions this evening :).

    Final Solution :

    (1/2mv^2 + mgh - Mk*mg*cos(theta)) / (mg*sing(theta)) = 40 meters

    ^^However, this is incorrect according to Mastering Physics.

    EDIT : Yes, the different velocities are where my confusion lies.

    **********************BELOW IS EPIC FAIL***********************************************
    **EDIT : I solved for v2 and plugged it into the original equation two posts ago, and got 9.74 meters instead of 40. HUGE difference; so either my lack of sleep is causing bad algebra or ......think its that one haha

    v2 = 9.9 m/s

    ^^(1/2mv2^2 - Mk*mg*cos(theta)) / (mg*sing(theta)) = 9.74 meters

    **********************DONT LAUGH, really tried at this point********************************
     
    Last edited: Nov 30, 2009
  11. Nov 30, 2009 #10
    I believe 40m should be he correct answer. If the textbook answer is 20, then I would assume the question is simply asking about the height the student goes up rather than the actual hypotenuse distance.

    Otherwise I don't think I'm missing anything from the problem. And I understand, fatigue is the worst. You can literally wake up in the morning and solve a problem you were stuck on for 2 hours last night in 10 mins, I think everyone gets moments like that.

    Edit: Yes, the fatigue is definitely affecting you. v2 should be bigger than v1, it would make sense the student is faster at the bottom of the hill than on top. Yet your v2 is less than v1. My calculation gives me 19.9m/s for v2, so it seems like you missed a 10 there. Also, it would help if you share the answer in the textbook so we can look at what might have went wrong.
     
  12. Nov 30, 2009 #11
    Unfortunately, I don't have the textbook answer because this is an even numbered problem (11.56). I have one more attempt on the problem, i might as well put 20 to see what happens :) . I really appreciate your help throughout both problems tonight and God Bless ya!

    EDIT: With my last attempt, i put 20 meters and according to mastering physics, that too is incorrect. I will email my instructor tonight to see if maybe Mastering Physics is incorrect or find out where i went wrong.

    Thanks again!
     
    Last edited: Nov 30, 2009
  13. Mar 30, 2011 #12
    I managed to get this one right (and all I'll say is that it is between 20 and 40 m !). I had the kinetic energy at the bottom of the hill divided by m*g*sin(theta)+(Mu of k)*m*g*cos(theta). What really saved me was the substitution of the height with the distance*sin(theta), which was suggested above.
     
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