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Simple work and accleration problem

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A rope exerts a constant horiztonal force of 250N to pull a 60kg crate across the floor. The velocity of the crate is observed to increase from 1m/s to 3m/s in a time of 2 seconds under the influence of this force and the frictional force exerted by the floor on the crate
    A) what is the accleration of the crate?
    b) what is the netforce acting on the crate?
    c) what is the magnitude of the frictional force acting on the crate?

    2. Relevant equations
    As stated in the attempt


    3. The attempt at a solution
    a) because of the given information i use the difference in velocity by time to produce a result of 1m/s. im not sure if i should be a = fnet/mass

    b) i use the eq force = ma, 60kg * 1m/s = 60N

    c) This is where i am having difficulty, i know for the accleration to be 1m/s, the equation would ultimatey have to be a = fnet/mass = 60n/60kg. subtracting 60n from the 250 yeilds the number, 190, which is the one i need to subtract from the given force to get 60. but i am having a hard time understanding how a 60N force on a crate can produce 130n in friction? Am i way off with my attempts?

    Thank you for your help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi highcontrast! Welcome to PF! :smile:

    (your a and b are fine, and no you don't need to use the mass to get the acceleration from the velocities … it's just a matter of geometry :wink:)
    why are you subtracting 60 twice? :confused:

    applied force 250 one way, friction force 190 the other way, so net force = 60, so acceleration = 60/60 = 1 …

    what's worrying you about that? :smile:
     
  4. Oct 20, 2009 #3
    That's a much more simple way of puting it! I think I may just be confused about the concept of 'net force' and not all that confident in my abilities. The answer definitely made sense in my head. The notion of 60N force acting on the box is what perplexed me. I, somehow, assumed that to be the netforce acting only on the box, but what I failed to realize is that its one system, meaning that 60N netforce would have been the result of the frictional force, and the pulling force.. Am I right in this line of thought?
     
  5. Oct 20, 2009 #4

    tiny-tim

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    Hi highcontrast! :wink:

    Yes, the net force (the total force) is all the external forces on the body,

    and it's always the F in F = ma.

    As you say, the box is "one system" …

    so in the end we can always lump all the forces together as "one force" (and similarly "one torque"). :smile:
     
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