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Homework Help: Find stopping distance from mass velocity & frictional force

  1. Feb 27, 2016 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    A 50,000kg locomotive is travelling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?



    t= 10/2.84*10^-3
    t=3521 sec

    d=35211 metres

    Stopping distance = 35.21 km? This sounds way too long. Can anyone point me in the right direction?
    Last edited by a moderator: Feb 28, 2016
  2. jcsd
  3. Feb 27, 2016 #2
    Not too far off.

    Is d the initial velocity times stopping time or average velocity times stopping time?
  4. Feb 27, 2016 #3
    d is the initial velocity times stopping time
  5. Feb 27, 2016 #4
    That is what you did, but is it right?
  6. Feb 27, 2016 #5
    A 50,000kg locomotive is travelling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?
    something wrong here ? check the third equation you are using
    Reference https://www.physicsforums.com/threa...e-from-mass-velocity-frictional-force.859569/
  7. Feb 27, 2016 #6
    Average velocity = displacement/time. Wouldn't the displacement be the same as the distance, as it is only travelling in a straight line? therfore yielding the same answer? Is there another way I can work out distance from what I have? Is what I have for time even correct? 3521 seconds or 58.6 minutes sounds too long as well.
  8. Feb 27, 2016 #7
    d=17605.65 metres

    Is that correct?
  9. Feb 28, 2016 #8
  10. Feb 28, 2016 #9
    ##d=\frac{a \cdot t^2}{2}+v_0 \cdot t##

    so ##d=\frac{v_0^2 \cdot m}{2 \cdot F}##

    ##d=\frac{10^2 \cdot 50000}{142}≈35211.2676##
    Last edited: Feb 28, 2016
  11. Feb 28, 2016 #10
    Now, this is a good equation to work off of, so I just want to say the KE is just 2,500,000Joules, KE=1/2(50,000kg)(10m/s)^2 that is a lot of work.
    We could assume working against a ƒ of 142Newtons, so I'll treat it in the same fashion as work against gravity, or work done against gravitational forces but we are working against a set force of ƒ. I could just take 2,500,000Joules/142N's of ƒ and get 17605.63 meters to slow to zero m/s......just some work-energy basics, as there is a force, a opposing force, we could use that but we would refer to is it as negative force doing negative work.

    However, lets pretend we have µ, a "mu" or a "moo" of 0.000289 and we don't know our Ffric, thus our ƒ is µ*m•g, so (.000289)(50000kg)(9.81m/s^2)= 142, I found this by taking the Fgrav/Force norm of 50000kg • 9.81m/s....which is 490500 Newtons, take 142N Friction force / 490500 Newtons of Fgrav/Fnorm(note that Fgrav and Fnorm are the same by Newtons 3rd law), we can find our µ or friction coefficient.

    We take 142N's Ffric/ 50000kg's to 0.00284 m/s^2 deceleration....

    now, find time we do t=Vf-Vi/a, so a=(0m/s)-(10m/s)/0.000289m/s^2 so we get 3521.12 Seconds.

    to find displacement, we can take this data to get our displacement or stopping distance, d=vi•t=1/2(at)

    d=(-10)(3521.1267s)+1/2(0.00284)(3521.1267s^2), we can get our final displacement 17605.63 but we would write that as -17605.63 as the -10m/s we are working against the opposing velocity of train as we are using Ffric to stop the train, a retarding or opposing force. It is written because that is the final DISPLACEMENT of the train with the force acting on it, that is why it is negative.
    Last edited: Feb 28, 2016
  12. Feb 28, 2016 #11
    ##d=\frac{10^2 \cdot 50000}{2\cdot142}≈17605.6338##
    I calculated wrongly.
  13. Feb 28, 2016 #12
    hehe, I was about to say dude
    Now there is another, if "better" way to calculate this using another equation.
  14. Feb 28, 2016 #13
    It would be much better if he knew some equations, specifically just displacement but he could check his work entirely with the equation of Vf^2=Vi^2 + 2ad, he would be finding the net force exerted on the train, so he could check his work with some dynamic problems.
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