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!

Kinetic Friction problem

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A small block of mass 1.74 kg rests on the left edge of a block of length 3.05 m and mass 7.95 kg. The coefficient of kinetic friction between the two blocks is 0.279, and the surface on which the 7.95 kg block rests is frictionless. A constant horizontal force of magnitude 11.4 N is applied to the 1.74 kg block, setting it in motion.

    How long will it take before this block reaches the right side of the 7.95 kg block? The acceleration of gravity is 9.8 m/s2 . (Note that both blocks are set in motion when ~ F is applied. Answer in units of s.


    2. Relevant equations
    Ffk=Uk(FN)
    X=(1/2)at^2
    F=ma


    3. The attempt at a solution
    so to find the normal force (FN) that mass 1.74 exerts on the block below it I did 1.74kg*9.8m/s^2 to get a normal force of 17.052N. I then found the kinetic friction by multiplying the normal force by the coefficient of kinetic friction: 17.052N*.279=4.757N. I then found the net force exerted on the top block to be 11.4N-4.757N= 6.64N. so using F=ma I used the numbers: 6.64N=1.74kg*(a) and solved for a to be a=3.8175m/s^2. Then I used X=(1/2)a(t^2): 3.05m=(1/2)(3.8175)(t^2) and got t=1.264s. However, this time is apparently not correct and I am confused as to what I am doing wrong. Do I need to take into account that the bottom block is moving aswell?
     
  2. jcsd
  3. Dec 4, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That would be a good idea. You have found the time it would take for the top block to reach the end of where the lower block was initially, but it has moved on.
     
  4. Dec 4, 2016 #3
    Thankyou!!!!!!!!! So I fixed my mistake by doing 3.05+X= (1/2)(3.817)(t^2) where x is the distance that the bottom block traveled and its equation is: x=(1/2)(.598427)(t^2) ( I got .598427 by doing 4.7575= 7.95a where 4.7575 is the friction force acting on the bottom block by the top block) and so I combined the two equations together and got T=1.3766seconds which turns out to be correct. Thank you for your tip!
     
  5. Dec 4, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Kinetic Friction problem
Loading...