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Newton's Methods

  1. Sep 26, 2005 #1
    I have 2 questions:

    1) A freight train has a mass of [tex]1.1\times 10^7 kg[/tex]. If the locomotive can exert a constant pull of [tex]7.8\times 10^5 N[/tex], how long does it take to increase the speed of the train from rest to [tex]90 km/h[/tex]?

    I figured I could just use the formula [tex]F=ma[/tex] to get the acceleration then I plugged that into [tex]v=at+v_0[/tex] to get the time. I came up with the answer of 1267.61. I might have a problem with units because [tex]a[/tex] was [tex]m/s^2[/tex] and [tex]v[/tex] was in [tex]km/h[/tex]. If thats the case how do I go about converting [tex]m/s^2[/tex] to [tex]km/h[/tex] or vice versa?

    2) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. Let the elevator have a mass of 1200 kg and an upward acceleration of 2.1 [tex]m/s^2[/tex]. Find [tex]T[/tex].

    With this one I just plugged it into the equation [tex]T=ma[/tex] but I dont know if this is the correct equation. I got an answer of 2520[tex]N[/tex].

    If anyone could help me with this please do.
     
  2. jcsd
  3. Sep 26, 2005 #2
    convert your km/h in the question to m/s in this way
    1km/hr = 1000m/3600secs

    why did you equate T with ma? Is tension the only force causing this acceleration? Keep in mind that the DIRECTION of these forces. Assume one direction to be positive and one to be negative and write down the force equation for this. That is
    ALL Net forces = Net Force = m x Net Acceleration
     
  4. Sep 26, 2005 #3
    I dont know where I got that equation from. I'm still a little confused about the tension problem. I guess I don't really know how to go about solving it.
     
  5. Sep 26, 2005 #4
    Ok. Draw a box taht represents the elevator. Assume up to be positive and down to be negative. What force points up, and what force points down? Now add these forces. But use the proper sign convention as in waht points up in positive and what pints down is negative. Now which way is the elevator accelratin? Is this acceleration (using the reference system) postive or negative. What is the force due to this acceleration? Now keep in mind that
    Sum of all the forces (poiting up and down) = Net Force(this may be positive or negative, depending on which way teh elevator is accelerating.
     
  6. Sep 26, 2005 #5
    Ok so what I get out of that is that I'm trying to find T which is the tension on the elevator pointing up? And the force down is the gravitational force which is 9.8 [tex]m/s[/tex]. The elevator is accelerating up at 2.1[tex]m/s^2[/tex]. So now I need to find an equation to figure out [tex]T[/tex].
     
    Last edited: Sep 26, 2005
  7. Sep 26, 2005 #6
    What is the formula for force? Thus how do you find the force exerted by gravity? ANd the net force?

    Also remember to make a REFERENCE SYSTEM. One should be positive and one be negative. Are Tensio nand gravity in the same direction? Correct the signs!
     
  8. Sep 26, 2005 #7
    [tex]F=ma[/tex]. Force up is possitive 2.1[tex]m/s^2[/tex] and force down is negative 9.8[tex]m/s^2[/tex]. So does that mean total [tex]F=[/tex]negative 7.7[tex]m/s^2[/tex]?
     
    Last edited: Sep 26, 2005
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