1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help Finding the Applied Force to move a box?

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A certain force moves a 4-kg box from rest to a velocity of 6 m/s in 3 seconds. If the coefficient of friction between the box and the floor is 0.05, what is the magnitude of the force?
    Given/Known:
    mass = 4 kg
    μ (I'm assuming it means kinetic friction?) = 0.05
    time = 3 seconds
    initial velocity = 0 m/s
    final velocity = 6 m/s
    gravity = 9.8 m/s squared


    2. Relevant equations
    Frictional Force = μ(mg)
    Net Force = (m)(a)
    final velocity = (initial velocity)(acceleration x time)


    3. The attempt at a solution
    From these equations, I calculated the acceleration to be 2 m/s squared. This is where I get confused - if you find the net force with the equation mass x acceleration, then you get 8 N for a net force. I calculated the frictional force with the equation above, so it was like this: (0.05)(4)(9.8) and I got an answer of 1.96 N. The natural force on the box is found by multiplying mass times the force of gravity, which ends up being 39.2 N. Now, my question: do I subtract the frictional force of 1.96 N (if that is right) from the natural force of 39.2 N to get the answer I'm looking for? Does the net force even get used in this problem? We only did a problem to find the acceleration with a given force, but not the other way around. Thanks to anyone that can help me!
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have to look in the horizontal and vertical directions separately. You have correctly solved for the acceleration and net force in the horizontal direction. So if the net force is 8 and the friction force is about 2, then the applied force must be what? Note that the net force is always in the direction of the acceleration , and that the friction force opposes the direction of relative motion. Note also that in the vertical direction, there is no acceleration.
     
  4. Apr 24, 2012 #3
    So, would I be correct in subtracting the friction force of 1.96 N from the net force of 8 N, since they oppose each other?
     
  5. Apr 24, 2012 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, the net force is the algebraic sum of the forces, that is, Fnet = Fapplied plus Ffriction, watch plus and minus signs .
     
  6. Apr 24, 2012 #5
    Ah, I think I understand. If Fnet = Fapplied plus Ffriction, and Ffriction opposed the Fnet making it negative, then that would make the equation Fnet = Fapplied minus Ffriction, so the Fapplied would be Fnet plus Ffriction...is that correct?
     
  7. Apr 24, 2012 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes!
     
  8. Apr 24, 2012 #7
    Thus making it 8 N plus 1.96 N, or 9.96 N...thanks a ton!
     
  9. Apr 24, 2012 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, good, but you should round off your answer to 10 N.
     
  10. Apr 24, 2012 #9
    Right, to account for significant figures..got it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help Finding the Applied Force to move a box?
Loading...