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How long an object takes to stop - using friction, momentum

  1. Apr 15, 2017 #1
    1. The problem statement, all variables and given/known data

    A block of mass 100 grammes was stationary on the flat surface at ##x = 0##. At
    time ##t = 0## a horizontal force of 10 newtons was applied on the block in the
    positive x direction during ##\Delta t = 3## seconds.

    Find at what x position the block will stop, if the coefficient of kinetic
    friction between the block and the surface is ##\mu_k = 0.2##.


    2. Relevant equations

    ##I = F \Delta t = m \Delta v = \Delta p## (where ##p## is used for momentum)

    3. The attempt at a solution


    I have that

    $$
    F = ma
    $$

    Then

    $$
    \frac{10}{0.1} = a
    $$


    And I can presume (?) that acceleration is constant.


    Then from this I have

    $$
    a = 100
    $$

    So if I have ##a = 100## then I can use Impulse as ##I = F \Delta t## to find

    $$
    F \Delta t = \Delta p
    $$

    So

    $$
    (0.1) (100) (3) = \Delta p = 30
    $$


    So from ##\Delta p = 30## I have ##m(v_f - v_0) = 30## and as ##v_0 = 0## this is
    ##mv_f = 30## and ##v_f = \frac{30}{0.1} = 300##

    This gives me the velocity as ##300## m/s at the point the force has been
    removed.

    From here I need to know how long it takes for the force of friction to slow the
    object to a stop.

    This means all of the energy from the point ##t = 3## is spent on friction.

    So

    $$
    \frac{1}{2} mv^2 = \text{work done against friction}
    $$


    So here I have that work done is ##W = F \Delta x## so given ##\mu_k = 0.2##, the
    mass of the object is ##0.1## then I would have the force of friction as

    $$
    F_{friction} = mg \times \mu_k = 0.1 \times 9.81 \times 0.2 = 10.11
    $$


    So then I have

    $$
    \frac{1}{2} (0.1) v^2 = 10.11 \Delta x
    $$

    And so

    $$
    \frac{1}{2} (0.1) v^2 \times \frac{1}{10.11} = \Delta x
    $$

    Using the velocity found from before this gives


    $$
    \frac{1}{2} (0.1) (300)^2 \times \frac{1}{10.11} = \Delta x
    $$

    So ##\Delta x \approx 445.103## meters.

    I don't really have much of a handle on what to expect from an answer here.

    I think what I've done is reasonable though?

    Thanks
     
  2. jcsd
  3. Apr 15, 2017 #2

    PeroK

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    You need units here. Otherwise, that looks right.

    Yes, but it would have been much easier just to multiply the acceleration by ##3## seconds!

    You should be able to see without a calculator that multiplying ##9.81## by two small fractions is not going to give you something bigger.

    Finally, you also need to think about how far the block moves during the acceleration phase.
     
  4. Apr 15, 2017 #3
    Ah yes >.< , whoops.

    So that the force of friction would be ##0.1 \times 9.81 \times 0.2 = 0.1962##.

    The means that during the acceleration phase (from ##t = 0 ## to 3) I have the net
    Force would be

    $$
    F_n = 10 - 0.1962 = 9.8038
    $$

    Given this I have the work done against friction over the three seconds as

    $$
    F_n \times \Delta x
    $$

    But I'm not sure how to find ##\Delta x## from this?

    Can I use the conservation of energy here?

    From that I would have that ##E = U + K + constant##. But I don't really have any
    potential energy, and the kinetic energy at the start is zero (so this seems
    like a poor choice).

    Oh - I can just use the equations of motion here then, so I would have

    $$
    x = x_0 + v_0 t + \frac{1}{2} a t^2
    $$

    So here this is just ##x = 0.5 \times 100 \times 9 = 450##. So that I would have
    moved ##350## meters during ##\Delta t## as a result of the force.

    But is that correct? Do these equations work when there's friction present?
     
  5. Apr 15, 2017 #4

    PeroK

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    Yes, of course, you have friction during the acceleration phase as well.

    I think it may be simpler to use equations of motion. There's no reason these won't work with friction, as long as you include the friction force when you calculate acceleration.

    What you know is:

    1) A force (10N - friction) acts for 3 seconds. The block accelerates, reaches a speed ##v##, and travels a distance ##d_1##, say.

    2) Friction slows the block from the speed ##v##. The block travels a further distance ##d_2## until it stops.

    You should be able to calculate the two distances from the equations of motion.
     
  6. Apr 15, 2017 #5
    So I have ##a = 100, v = 300## after three seconds.

    Then I remove the force that's pushing the object and friction slows it to a stop.

    Then at three seconds I have the kinetic energy

    $$
    \frac{1}{2} m v^2
    $$

    And this is all spent on work against friction (so that the object comes to a
    stop ), giving

    $$
    F \Delta x = \frac{1}{2} m v^2
    $$

    So I have ##F = m g \mu_k## then

    $$
    \Delta x = \frac{\frac{1}{2} v^2 }{g \mu_k}
    $$

    Which is

    $$
    \Delta x = \frac{0.5 \times 300^2}{9.81 \times 0.2} \approx 22.935
    $$

    So that I travel around 22.8 km it seems! (really? )

    Is this right? The approach etc.

    Thanks
     
  7. Apr 15, 2017 #6

    PeroK

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    You're forgetting that the friction force is acting during the acceleration.

    Is that ##22.9m##?
     
    Last edited: Apr 15, 2017
  8. Apr 15, 2017 #7

    PeroK

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    I did a quick calculation using ##g = 10ms^{-2}## and got exactly ##1.8 km##. That includes the distance accelerating and decelerating.
     
  9. Apr 15, 2017 #8
    Sorry - that answer is correct but Its in thousands. I'd be unlikely to trust
    any boss who gave me a job anyway tbh :)


    I thought that I had accounted for the frictional force during that, damn.

    OK

    $$
    \text{Force net} = 10 - mg\mu_k
    $$

    Then we have that

    $$
    F_n = ma
    $$

    So

    $$
    a = \frac{F_n}{m}
    $$

    From this we can use impulse as

    $$
    I = F \Delta t = (10 - mg \mu_k) \times \Delta t = \Delta p
    $$

    So rearranging to solve for ##v_f## , where ##v_f## is the velocity at ##t = 3##
    seconds gives

    $$
    \frac{
    (10 - mg \mu_k) \times \Delta t
    }{
    m
    } = v_f
    $$

    We can use this value of ##v_f## to find the kinetic energy at ##t = 3## seconds

    $$
    E = \frac{1}{2} m v_f^2
    $$


    We know that all this energy is spent on work to bring the object to a stop.

    Work ## = F \times \Delta x##, where ##F = mg \mu_k## so we have

    $$
    \Delta x \times mg\mu_k = \frac{1}{2} m v_f^2
    $$

    Dividing by ##mg \mu_k## gives

    $$
    \Delta x = \frac{\frac{1}{2} m v_f^2}{mg \mu_k}
    $$

    So we have to plug values here, velocity at 3 seconds is

    $$
    v_f = \frac{10 - 0.1 \times 9.81 \times 3}{0.1} = 294.114
    $$

    We have that the kinetic energy is equal to the work done which is ##F\Delta x##,
    where here ##F## is just due to friction and is ##F = mg \mu_k = 0.1 \times 9.81
    \times 0.2 = 0.1962 ##

    Then we can sub this in to the expression for ##\Delta x## as

    $$
    \Delta x = \frac{\frac{1}{2} (0.1)(294.114)^2}{0.1962} \approx 22044.60882
    $$

    So this is ##\approx 22## km

    does that make sense?

    thanks
     
  10. Apr 15, 2017 #9

    PeroK

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    My apologies, I had put a mass of ##1kg## in my "quick" calculation. I was trying to be too clever.

    You're still forgetting the distance it travels during the acceleration.
     
  11. Apr 15, 2017 #10

    PeroK

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    As an aside, let me show you a neat trick for these sorts of problems. You may or may not wish to remember this.

    If an object accelerates from rest at a constant acceleration, then decelerates from this speed to rest at a constant deceleration. Then the average speed during both legs of the journey is the same: it's half the maximum speed.

    You could solve this problem by working out the maximum speed, the time for each leg of the journey and the distance travelled would be half the maximum speed times the total journey time.
     
  12. Apr 15, 2017 #11
    Hrm ok - i'll think about how to apply that to this (i'm not sure on first read)

    is the rest of the problem sound now though?

    thanks
     
  13. Apr 15, 2017 #12

    PeroK

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    The working is correct, but you definitely need units. You'll lose marks for that in an exam.
     
  14. Apr 15, 2017 #13
    OK cool - thanks for the help
     
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