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Homework Help: Force and Motion, horizontal forces

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data
    20. A 873-kg (1930 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9mph) in 0.59 s.
    -----a. Find the average acceleration of the dragster during this time interval.
    My solution:
    Given variables:
    M=873kg
    Vo=0
    Starting time: 0
    V=26.3m/s
    T=0.59
    Formula to be used: V=Vo + at converted into A=v-vo/t
    It equals:
    a= 26.3/0.59=44.6m/s^2
    -----b. What is the magnitude of the average net force on the dragster during this time?
    My solution:
    Given variables:
    m=873 kg
    a=44.6m/s^2
    F net= ma
    F net = 38,935.8 N
    F net = 39,900N
    -----c. Assume that the driver has a mass of 68kg. What horizontal force does the seat exhert on the driver?
    Here is where i have a problem..[B/]
    I would assume I have to do:
    F net=ma using 68kg as my M and 44.6m/s as my A to solve for the force exherted by the seat, which would equal 1,788.4N (Horizontal Force seat exherts) Yet I dont know if thats done correctly.. I would love someone to revise over the work, especially C to see if I did it correctly..
     
    Last edited: Feb 27, 2007
  2. jcsd
  3. Feb 27, 2007 #2
    Looks good to me.
     
  4. Feb 27, 2007 #3
    Awesome,

    Well thats great, then I guess that means im doing the rest of the homework right :P! Hehe thanks
     
  5. Feb 27, 2007 #4
    Help again i think

    I think im stuck again,

    37. a sled of masss 50.0kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10
    -----a. What does the sled weigh?
    Im assuming they mean force, so it would be 50kg x 9.8 to convert to N,so 490N, and get its "weight"? I think this is how it is, although it seems too easy, if anyone would revise this id be grateful :)
    -----b. What force will be needed to start the sled moving?
    My given values are:
    m=50kg
    us(coefficient of static friction)=0.3
    uk(coefficient of kinetic friction)=0.1
    Fp=uk*m*g which is Fp=0.1*50*9.8= 49N
    So is the force needed to start 49 newtons?
     
    Last edited: Feb 27, 2007
  6. Feb 27, 2007 #5
    looks good so far
     
  7. Feb 27, 2007 #6
    I think the "So is the force needed to start 49 newtons?" is wrong, because now that I think of it wouldnt I insttead use the Fs= us x mg which would equal 147 N (which is 0.3 x 50 x 9.8)..
     
  8. Feb 27, 2007 #7
    You need to overcome static friction to get the object to move.
     
  9. Feb 27, 2007 #8
    oh so then its like.. above 147? like 148+? Cause static friction is 147N, so yeah, I guess 148N would be overcoming it

    thanks again for the fast reply
     
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