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Having trouble solving Nonconservative Force problem

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A 0.75-kg block slides on a frictionless surface with a speed of 20 m/s. It then slides over a rough area 1.0m in length and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17.What is the speed of the block after it passes across the rough surface?

    m = 0.75 kg
    x(during friction being applied)= 1.0m
    Coefficient of Friction = 0.17
    Initial Velocity = 20 m/s
    Gravity = 9.8​



    2. Relevant equations

    -Equations I used -
    () - means multiplying.

    Normal Force = Mass(Gravity)
    -Just used the Fx component since Fy equals 0-
    =
    Friction Force= Coefficient(Normal Force)(Distanceof the friction)

    K = 1/2mv2

    Ei = Ef
    =
    Ki - Friction Force = Kf
    =
    1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2



    3. The attempt at a solution
    I first realized what I needed to find, which was the Final Velocity.

    Then set up my equation.
    1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

    Then plugged in the numbers.
    1/2(.75kg)(20m/s)-(.17f)(.75kg)(9.8m/s2)(1.0m) = 1/2(.75kg)(v2)

    Simplified.
    6.2505 = 3.75 v2

    Simplified again.
    1.6668 = v2

    Squared.
    1.291 = v

    Books Answer
    However, When I looked at the book they got the answer 0.82m/s and got it by doing...
    v=√(InitialV2) - 2 (Coefficient of friction)(Gravity)(Distance))
    * Note - all of this is under the square root *
    v = √(2.0)2-2(0.17)(9.8)(1.0)
    =0.82 m/s

    My question is... What did I do wrong and what did the book do correctly? I would really love a good explanation on this as It's important for me to grasp this concept. The answer is there but I want to know HOW and WHY. Thank you in advance whoever takes the time to answer this it is greatly appreciated and your positivity will be rewarded.

    Kind regards, A lost soul.
     
  2. jcsd
  3. Oct 11, 2013 #2

    PhanthomJay

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    You used energy methods nicely, but made a math error by not squaring the 20...check your work again and you should see it should have been K_i = 1/2 m (20)^2. The book used the kinematic equation after first solving for the acceleration during the friction phase. Either way is ok.
     
  4. Oct 11, 2013 #3

    gneill

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    Hi Iconic, Welcome to Physics Forums.

    The usual convention is to place the x-axis horizontal and the y-axis vertical. Thus the normal force points along the y-axis, normal (perpendicular) to the Earth's surface (which is why it's called the "normal force").

    Nooo. Force is not work. When you multiply a force by a distance you get work done by that force moving something through that distance. The friction coefficient multiplied by the normal force yields the friction force. ##\mu_k F_n = F_f##.

    Again, force is not energy. You want an energy term there. In particular you want the work done by the Friction Force on the mass m over the 1.0 meter distance.

    That last line is good, except you should use different variables for the initial and final velocities.

    That value is highly suspicious. The formula is fine but the quoted result is obviously wrong. Maybe a typo in the book? Consider that the block starts out going 20 m/s and passes over a mere 1.0 meter of barely rough surface (0.17 as a friction coefficient is pretty small). I'd expect a value much closer to the initial speed.

    EDIT: Re-reading the numbers through your post I see that you meant the initial speed to be 2.0 m/s, not 20 m/s! Big difference! The book's answer is correct. It' really is easier to follow symbols than numbers with decimals when parsing another's work.

    It pays to leave the problem in symbolic form as long as possible. During your workings you wrote:

    1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

    Let me translate that into slightly more compact form, writing out the normal force as it was formulated:
    $$\frac{1}{2} m v_i^2 - (\mu_k m g) d = \frac{1}{2} m v_f^2$$
    What do you get if you multiply though by 2 and divide through by m? Look familiar?
     
    Last edited: Oct 11, 2013
  5. Oct 11, 2013 #4
    I see what you are saying however, after realizing my mistake I now tried to solve the equation with this in mind and still ended up with an incorrect result.

    1/2(.75kg)(20m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = 1/2(.75kg)(v2)

    Simplified.
    150-1.2495 = .375(v2)
    Simplified Again.
    148.7505 = .375(v2)
    Step 3
    396.668 = v2
    Step 4
    √(396.668) = v
    Step 5
    19.91 = v

    Still not right =/
     
  6. Oct 11, 2013 #5

    gneill

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    Iconic: Check the edit that I'd added to my previous post.

    19.92 m/s (rounded) would be correct if the initial speed was 20 m/s :smile:
     
  7. Oct 11, 2013 #6
    Sorry, for replying so late I'm just taking my time with these posts to ensure I didn't mess up anywhere. Just read your post gneill, THANK YOU very much for that broad and well written explanation. Couple things I need to say though, there might have been a typo, however; after double checking it still says 20 m/s in the question and 2.0 when the book solves for Vf:bugeye:. But anyways, when you multiply 2 from 1/2mvf2 on to the other side and divide out m is this how the equation should look?

    1/2mvi2-Coefficient(Normal Force)(Distance)= 1/2mv2
    *2
    mvi2-Coefficient(Normal Force)(Distance)= mv2
    /m
    vi2-Coefficient(Normal Force)(Distance)= v2

    Solving: Assuming Vi was 20

    (20m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = v2

    400 - 1.2495 = v2

    398.7505 = v2

    √(398.7505)

    v=19.96

    Solving: Assuming Vi was 2.0

    (2.0m/s2)-(.17f)(.75kg)(9.8m/s2)(1.0m) = v2

    4 - 1.2495 = v2

    2.7505 = v2

    √(2.7505)

    v= 1.658

    Assuming this was a typo... I'm not getting the right answer =/ Was my algebra incorrect? I feel like I'm not manipulating the equation correctly.:frown:

    EDIT - wanted to point out that I realize this equation is supposed to look similar if not exactly like the kinematic equation
    vf2 = vi2 + 2a Delta X
    But I'm horrid with algebra and am having a difficult time getting it there.
     
    Last edited: Oct 11, 2013
  8. Oct 11, 2013 #7

    gneill

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    Ooops! You forgot the middle term. When you multiply through or divide through an entire equation you must include every term.

    [SNIP --- the above affected the excised lines]
    :
    :
    :
    Yup. You missed multiplying the friction work term by two. Give it another go.

    Practice will polish your skills. Everyone climbs the same mountain :smile:
     
  9. Oct 11, 2013 #8
    So what I really do is...

    1/2mv2-Coefficient(Normal Force)(Distance)= 1/2mv2

    Multiply the two from all the variables on the opposing side? Not just 1

    mv2- 2Coefficient(Normal Force)(Distance)= mv2

    Then Mass is divided canceling out every where

    v2- 2Coefficient(Normal Force)(Distance)= v2

    This makes sense now. Only 1 question and I know this is a dumb one but I might as well ask:redface:. How come Normal Force and Distance isn't multiplied by 2 as well? Is that because that's supposed to equal Work of Friction so it's essentially 1 variable?
     
  10. Oct 11, 2013 #9

    PhanthomJay

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    You are still having algebra problems. You forgot to divide the middle term by m when you divided the other terms by m. The normal force is mg. the friction force is (u)mg . The magnitude of the work done by friction is (u)mgd. Be sure to divide that term by m also, and the equation will be exactly the same as the kinematic equation, as it should be.
     
  11. Oct 11, 2013 #10

    Ahh, thank you for clearing that up I was having difficulty with defining some forces but that explains it. So it's not the friction force we were trying to get but the magnitude of that frictional force over 1.0m. Thank you so much that really helped!:approve:

    v2- 2(u)gd= v2
     
  12. Oct 11, 2013 #11

    PhanthomJay

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    Iconic

    You said
    when you meant to say
    . You are given vi = 2.0 m/s, and you are trying to solve for vf.

    The proper way to do the algebra is to isolate vf to get the equation
    [itex]v_f = \sqrt{v_i^2 -2ugd}[/itex], from which
    [itex]v_f = \sqrt{4 - 2(.17)(9.8)(1.0)}[/itex] or thus
    [itex]v_f = \sqrt{4 - 3.332}[/itex]
    [itex]v_f = \sqrt {.668} = 0.82 m/s[/itex]

    Now if you find the algebra difficult, you can instead 'plug and chug':

    [itex]v_i^2 -2ugd = v_f^2[/itex]
    [itex]4 - 3.332 = v_f^2[/itex]
    [itex]0.668 =v_f^2[/itex]
    [itex]v_f = 0.82 m/s[/itex]

    This latter method is subject to errors of its own, and loses some information, but sometimes it makes the algebraic manipulations easier, in a pinch.
     
  13. Oct 11, 2013 #12
    Oops, I forgot to add the subscript i and f on that last equation but what you said was exactly what I meant. Again thanks, I think the only real issue I had with this problem was figuring out how we went from

    1/2mvi2-2(u)mgd=1/2mvf2

    to

    vi2- 2(u)gd= vf2

    But I understand now that you're supposed to manipulate the equation for the variable you want then solve. I didn't know that you had to multiply the 2 and cancel out the M in not just one but all variables- that's what the issue was and I get it now.:smile:
     
  14. Oct 12, 2013 #13

    PhanthomJay

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    I think you understand the algebra now, but note again your typo: You wrote
    that red 2 was not in the original equation , you meant to write
    by multiplying all terms by 2 and then dividing all terms by m. Just a typo. Good work.
     
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