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!

Energy and Work

  1. Apr 11, 2008 #1
    A 1000.0 kg car experiences a net force of 9500N while decelerating from 30.0m/s to 23.4m/s. How far does the car travel while slowing down?

    I do not even know how to start to setup the problem. I know that W=Fd but I do not know how to apply this with the velocities.

    Any help would be appreciated. Thank you.

    Stephen
     
  2. jcsd
  3. Apr 11, 2008 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can calculate the car's acceleration from the given data using Newton's laws, then use the standard kinematic equations of motion to calculate the distance.
     
  4. Apr 11, 2008 #3
    since we are on the energy and momentum chapter is there any way to do it with KE and PE?
     
  5. Apr 11, 2008 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, sure. Use the work energy theorem
    [tex]W_{total} = W_{net} = \Delta KE[/tex]. In general , you've got to be a bit careful when using this equation, because W_net includes work done by both conservative (like gravity) and non-conservative (friction, etc.) forces, but it this case, those forces are given as one net number, so you don't have to worry about PE change.
     
    Last edited: Apr 11, 2008
  6. Apr 11, 2008 #5
    so...
    Fd=1/2mvf^2 - 1/2mvi^2

    ?????
     
  7. Apr 11, 2008 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    yes, where the value you use for F is the given F_net. Check it out both ways using the energy method vs. Newton 2 and the kinematic equations.
     
  8. Apr 12, 2008 #7
    Would vi be 23.4m/s
    and vf= 30m/s
    ???

    If it is the other way around then F*D= a negative number

    any help would be appreciated.
     
  9. Apr 13, 2008 #8
    vf is 23.4m/s since you're slowing down, right? so the change in kinetic you get should be a negative number, and keep in mind that the force of friction is always negative, so the work done is negative too.
     
  10. Apr 13, 2008 #9
    so vi= 30.0m/s
    vf=23.4m/s

    and the force of 9500N is negative???

    -9500N*d=(1/2)(1000kg)(23.4m/s)^2 - (1/2)(1000kg)(30m/s)^2

    ???
     
  11. Apr 13, 2008 #10
  12. Apr 13, 2008 #11
    d= 18.5m or 19m with sig figs.?
     
  13. Apr 13, 2008 #12
    A general rule of thumb is to use the least amount of sig figs as the values given to you in the question.
     
  14. Apr 13, 2008 #13
    which would be 2 from 9500N
     
  15. Apr 13, 2008 #14
    do i have the right number of sig figs
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Energy and Work
  1. Energy and Work (Replies: 1)

  2. Calculate the lift work? (Replies: 15)

Loading...