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!

Change in Kinetic Energy/ Work

  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data
    000000000.PNG

    2. Relevant equations
    ΔKE=W+FcosΘd
    KE=.5mv^2
    Final Vel^2=initial Velocity^2+2ax

    3. The attempt at a solution
    So calculate the force at x=2.6 to -5.356N, set 2.6(-5.356)=1.3Vf^2-1.3(0^2) which seemed to prove a problem as the square of the final velocity was negative, but just used that negative square in the difference of kinetic energy equation and got -5.356 J- 0= -5.356 as the change in Kinetic energy or work. I am skeptical of my methods and also a little clueless on the second part of the problem.

    Many thanks
     
  2. jcsd
  3. Oct 8, 2016 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Please display the problem statement in a form that can be read with reasonable effort.

    Perhaps, type it out.
     
  4. Oct 8, 2016 #3
    A 2.6 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529127entrance1_N10035.mml?size=14&ver=1475946606896&algorithm=1&rnd=1475946607102 [Broken], where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.6 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.6 m?
     
    Last edited by a moderator: May 8, 2017
  5. Oct 8, 2016 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That's better. We also have:
    F(x) = (4.7 - x2)i newtons.
    or
    ##\ \vec F (x)=(4.7-x^2)\hat i \ ##​

    What is F at x = 0? What is F at other locations?
     
    Last edited by a moderator: May 8, 2017
  6. Oct 8, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your formula ##V_f^2 = V_i^2 + 2ax## does not work; it assumes that ##a## is a constant, which is not true in this problem.
     
  7. Oct 9, 2016 #6
    Okay so I integrated the force to 4.7x-x^3/3. I am unsure of what to do next
     
  8. Oct 9, 2016 #7
    I am silly. I got 6.36 J for part 1. Now part 2 is a bit more confusing
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted