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Newton's first and secon law

  1. Mar 4, 2004 #1
    A particle of mass m is traveling at an initial speed v0=25.0m/s.
    When a net force of 15.0N acts on it, it comes to a stop in a distance of 62.5m. What is m?

    I dont know how to get the accelration since I dont have the time t, so I can calculate the mass.

    May you please help me out?
  2. jcsd
  3. Mar 4, 2004 #2
    Answer... Hopefully

    Alright let me see if I can help you out at all

    well since V^2=Vo^2+2ax

    Therefore 0=25^2+2a62.5
    Therefore -625=2a62.5
    Therefore a=-5m/s^2

    Then I think you can do


    m=3 kg then I belive.
  4. Mar 5, 2004 #3


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    Staff Emeritus
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    Here's another way to do it, admittedly harder because it doesn't assume that formula, V^2=Vo^2+2ax, Tom McCurdy gave you.

    The force is -15N so, with mass m, the acceleration is a= -15/m.
    (I'm taking the force to be negative since it stops the motion.)

    The initial speed is 25 so, at any time t, the speed is
    25- (15/m)t.

    The distance moved in time t is 25t- (15/2m)t2.

    When the object stops, the speed is 25- (15/m)t= 0 and the distance is 25t- (15/2m)t2= 62.5.

    You now have two equations to solve for m and t. Since you were specifically asked for m, it might be simplest to rewrite
    25-(15/m)t= 0 as (15/m)t= 25 so t= (25/15)m= (5/3)m. Now put that into the second equation: 25(5/3)m- (15/2m)(25/9)m2= 62.5 which is the same as (125/3)m- (125/6)m= (125/6)m= 62.5 kg.
  5. Mar 5, 2004 #4
    Let me tell you the key to answering problems:

    1. eat a hogie

    2. do a little dance

    3. sing some Van Halen

    4. answer the prob

    works for me every time
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