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!

Physics problem

  1. Sep 15, 2004 #1
    A 1.50 * 10^2 N force is pulling a 50.0 kg box along a horizontal surface. The force acts at an angle of 25.0°. If this force acts through a displacement of 12.0 m, and the coefficient of friction in 0.250, what is the speed of the box, assuming it started from rest?
     
  2. jcsd
  3. Sep 15, 2004 #2

    ahrkron

    User Avatar
    Staff Emeritus
    Gold Member

    What have you done so far?
     
  4. Sep 15, 2004 #3
    this is how i attempted to solve the question:

    Fg=mg
    =(50kg)(-9.81m/s2)
    = -491 N

    Fnet=Fn-Fg
    Fn= 491 N

    Ff=uFn
    =(0.250)(491N)
    =123 N

    W=Ek
    W=Fd
    W=(123 N)(12.0 m)
    W=1472 J

    Ek=1/2mv2
    V=squareroot 2Ek/m
    V=squareroot 2(1472 J)/(50.0 kg)
    V= 7.7 m/s

    The actual answer is 2.53 m/s. Can you tell me what i am doing wrong, thanks.
     
  5. Sep 16, 2004 #4
    You have not accounted for the angle anywhere.
    Resolve the applied force into its horizontal and vertical component.

    -- AI
     
  6. Sep 16, 2004 #5

    cronxeh

    User Avatar
    Gold Member

    Ffriction=mu * Fn * cos (angle) -- this is the force due to friction, acting in opposite direction to pull

    ( pull <-- object --> Ffriction)
     
    Last edited: Sep 16, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Physics problem
  1. A problem (Replies: 24)

  2. A problem (Replies: 15)

Loading...