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  1. #1
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    Mechanics Help!

    I got 225000N for this question but my teacher's marking (which I'm not convinced with says 25000N - he's probably right though but I must check as I can not ask him as its the christmas holidays and our exam is pretty much the first day back)

    See what you make of it! Please =P

    A train of mass 200000kg is moving at a constant velocity of 25m/s, it then brakes and has constant decelleration and comes to rest in 40s. In this stage there is no driving force produced by the train's engine, but a retarding force produced by the train's brakes. There is also a constant resistance force on the train of magnitude 100000N throughout the brakin. Find the retarding force produced by the breaks.


    Thanks guys!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by anthmoo View Post
    I got 225000N for this question but my teacher's marking (which I'm not convinced with says 25000N - he's probably right though but I must check as I can not ask him as its the christmas holidays and our exam is pretty much the first day back)

    See what you make of it! Please =P

    A train of mass 200000kg is moving at a constant velocity of 25m/s, it then brakes and has constant decelleration and comes to rest in 40s. In this stage there is no driving force produced by the train's engine, but a retarding force produced by the train's brakes. There is also a constant resistance force on the train of magnitude 100000N throughout the brakin. Find the retarding force produced by the breaks.


    Thanks guys!
    I agree with your teacher. The impluse-momentum theorem says:
    I = F_{ave} \Delta t = \Delta p

    Now, since both the retarding force of 100000 N and the braking forces are constant, we may simply take F_{ave} = f + 100000 \, N (where f is the force due to the brakes) as a constant. Thus:
    -(100000 + f) \Delta t = m(v - v_0) = -m v_0 (The final speed of the train is v = 0 m/s.)

    So we have:
    -(100000 + f) \cdot 40 = - 200000 \cdot 25

    100000 + f = 5000 \cdot 25 = 125000

    f = 125000 - 100000 = 25000 \, N

    -Dan
    Last edited by topsquark; December 30th 2006 at 05:20 AM. Reason: Typo!
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    Quote Originally Posted by topsquark View Post
    I agree with your teacher. The impluse-momentum theorem says:
    I = F_{ave} \Delta t = \Delta p

    Now, since both the retarding force of N and the braking forces are constant, we may simply take F_{ave} = f + \, N (where f is the force due to the brakes) as a constant. Thus:
    -(100000 + f) \Delta t = m(v - v_0) = -m v_0 (The final speed of the train is v = 0 m/s.)

    So we have:
    -(100000 + f) \cdot 40 = - \cdot 25

    100000 + f = 5000 \cdot 25 =

    f = = = 25000 \, N

    -Dan

    I can follow you most of the way but I do not understand where the  5000 \cdot 25 came from! Where did it come from???


    PS. We weren't taught this theorum in class but it seems like a good one to know for the exam as it seems to make the question easier to answer
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    Quote Originally Posted by anthmoo View Post
    I got 225000N for this question but my teacher's marking (which I'm not convinced with says 25000N - he's probably right though but I must check as I can not ask him as its the christmas holidays and our exam is pretty much the first day back)

    See what you make of it! Please =P

    A train of mass 200000kg is moving at a constant velocity of 25m/s, it then brakes and has constant decelleration and comes to rest in 40s. In this stage there is no driving force produced by the train's engine, but a retarding force produced by the train's brakes. There is also a constant resistance force on the train of magnitude 100000N throughout the brakin. Find the retarding force produced by the breaks.


    Thanks guys!
    Here is one way, via F = ma, which may mean if there is an acceleration, then there must be a force that is causing it.

    F = force...in N in your problem here.
    m = mass, in kg here.
    a = acceleration, in m/sec/sec here.

    The accelaration here is actually deceleration, or acceleration to slow down, or "negative" acceleretion, or acceleration opposite in direction to the initial direction of the motion.

    a = (final velocity minus initial velocity) / time
    a = (0 -25)/40
    a = -(5/8) m/sec/sec

    The force causing this "negative" acceleration is the constant 100,000N plus the constant braking force, X, on the brakes, so,
    F = ma
    (100,000 +X) = (200,000)(5/8)
    100,000 +X = 125,000
    X = 25,000 N ..................answer.

    Why is the acceleation not negative in the equation?
    Because we take the F as positive. F and "a" here have the same direction. They are both opposite the direction of the motion of the train. If F is positive, so is "a".
    Last edited by ticbol; December 29th 2006 at 02:19 PM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by anthmoo View Post
    I can follow you most of the way but I do not understand where the  5000 \cdot 25 came from! Where did it come from???
    Let me insert an extra step:

    -(100000 + f) \cdot 40 = - 200000 \cdot 25

    \frac{-(100000 + f) \cdot 40}{40} = \frac{- 200000 \cdot 25}{40} = - 5000 \cdot 25

    100000 + f = 5000 \cdot 25 = 125000

    f = 125000 - 100000 = 25000 \, N

    Quote Originally Posted by anthmoo View Post
    PS. We weren't taught this theorum in class but it seems like a good one to know for the exam as it seems to make the question easier to answer
    Sorry about that. In that case you might wish to use ticbol's solution. The origin of the Impulse - Momentum theorem is easy enough, it's merely Newton's Second Law written sideways. For technical reasons I am going to let \sum F = F_{ave}, the average net force over a given time span acting on the system. (You can look up the particulars in your Physics book.)
    F_{ave} = ma = m \frac{\Delta v}{\Delta t}

    F_{ave} \Delta t = m \Delta v

    In a situation where the mass of the system is a constant, the RHS is simply the change in momentum of the system \Delta p. We define the LHS to be a quantity called the "impulse" (I), which is a kind of a "total push" we give to a system. Though I didn't use the symbols, note that the impulse is a vector just like the change in momentum. (If you are aware of the fact that Newton's Second Law is really \sum F = \frac{\Delta p}{\Delta t}, this derivation is ridiculously simple.)

    -Dan
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