1. Not finding help here? Sign up for a free 30min 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!

Frictional forces

  1. Feb 17, 2008 #1
    Frictional forces [SOLVED]

    1. The problem statement, all variables and given/known data
    A stockroom worker pushes a box with mass 11,2 kilograms on a horizontal surface with a constant speed of 3.5 m/s. The coeffisient of kinetic fricion between the box and the surface is 0.2.

    In a) i had to find the force that the worker must apply to maintain the motion.

    b) If the force calculated in part a) is removed, how far does the box slide before coming to rest.

    2. Relevant equations

    3. The attempt at a solution
    I am pretty sure that I solved task a). I found the frictionforce to be 22N, and since the box is having a constant speed, the sum of forces in x-direction have to be zero.
    Force = Frictionforce in this case.

    But i really need some tips on the B part. Anyone?

    Last edited: Feb 17, 2008
  2. jcsd
  3. Feb 17, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Once you remove the applied force of the push, the net force is no longer zero: You have the unbalanced friction force acting to slow the box down. Find the acceleration and use kinematics to find the distance. Or use energy methods.
  4. Feb 17, 2008 #3
    Well the force that was being applied only acted to keep the box moving at 3.5m/s so when it is removed the box will start deccelerating from this initial speed.

    You need to find the acceleration of the box due to the frictional force, which is dependant on the boxes mass. With this you can determine the distance the box travels.
  5. Feb 17, 2008 #4
    Thanks for fast replies:) Ok, I tried something like that, but I didn't come out with an correct answer. When the force is removed I have:

    Force-Friction = ma

    0 - Friction = ma

    a = Friction / m = -1.96m/s^2

    But how do I find the distance now?
  6. Feb 17, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    It's a kinematics problem now. Accelerated motion.
  7. Feb 17, 2008 #6

    I did of course make a mess of things with the sign. Got the distance to be 3,1 meters after finding the time 1.78s.
  8. Feb 17, 2008 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Sounds good.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Frictional forces
  1. Frictional Forces (Replies: 1)

  2. Friction Force (Replies: 3)

  3. Frictional Force (Replies: 6)

  4. Friction and forces? (Replies: 6)

  5. Force and Friction (Replies: 4)