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 Energy Problem

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A 5.0 kg mass is released from rest, at the top of a frictionless 30° inclined plane. Calculate the kinetic energy of the mass when it has slid 40cm down the incline starting from rest. (ANS = 9.80 J)


    2. Relevant equations
    KE=1/2(mv^2)
    a=v/t
    d=v1t + 1/2a(t^2)
    v2^2=v1^2 + 2ad
    v2=v1 + at
    d= (v1+v2/2)*t


    3. The attempt at a solution
    G: v1= 0 ; θ= 30° ; d=40cm
    R: KE=? ; v2=? ; t=?
    (NE direction is +ve)

    (1) v2y^2=v1y^2 - 2aydy
    v2y^2= √0 -2(9.80)(-0.40sin30)
    v2y= 1.98 m/s

    (2) v2y=v1y +at
    t = v2y/ay
    t = 0.202 s

    (3) v2x= dx/t
    v2x= 0.40cos30/0.202
    v2x= 1.71 m/s

    (4) v2= √(1.98)^2 + (1.71)^2
    v2= 2.62 m/s

    (5) KE = 1/2(mv^2)
    KE = 1/2(5.0*2.62^2)
    KE = 17.2 J

    My answer is far off from the correct one so i'm assuming my approach might be wrong.. But I just don't know from where it is that i'm doing it wrong.. Probably something with my calculations, not too sure entirely..
     
  2. jcsd
  3. Dec 15, 2013 #2

    CAF123

    User Avatar
    Gold Member

    The acceleration down the slope is not 9.8m/s2, what is it?
     
  4. Dec 15, 2013 #3
    not sure if it's zero or 9.8sin30
     
    Last edited: Dec 15, 2013
  5. Dec 15, 2013 #4
    would it be zero?
     
  6. Dec 15, 2013 #5
    The acceleration I believe would be Fnet=ma (in the x direction). So add up all the forces in the X direction, which seems to be only one because the surface is frictionless. So mgsin30=5a
     
  7. Dec 15, 2013 #6

    CAF123

    User Avatar
    Gold Member

    brown20b,
    Sorry, I misinterpreted your first approach - you have your axes horizontally and vertically rather than perpendicular and parallel to the slope which is the usual orientation of the frame for solving such a problem. But we can still do it your way if you wish.

    The axes are usually oriented such that one axis points in the direction of the net acceleration of the block, in this case down the slope and the other axis perpendicular to it. With this orientation, we arrive at the equation posted by azhang40 in one step.

    In your attempt, you assumed the acceleration in your y direction was g and the acceleration in your x direction was 0 It is not the case, since the resultant is then g and clearly the block is not in free fall.

    I am happy to go through the method with your x and y horizontally and vertically, or would you prefer to consider the reorientation of axes? Both methods follow the same process, one just takes a little longer.
     
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 Energy Problem
Loading...