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!

Work Energy Concept

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Ok im all confused on this one.

    ok so leave part (I) and (II)

    onto part (III)

    word done by the pulling force acting on the block = the loss of energy + work done by resisting force

    by loss of energy i mean gain in PE - loss in KE

    =(7500+225)-(1225)+(7.5 x 200) = 8000J

    now im confused as to whether or not the value of theta is mgsintheta or fscostheta please help
  2. jcsd
  3. Jan 27, 2007 #2


    User Avatar
    Homework Helper

    Don't overcomplicate things. The work done by all the forces acting of the block from A to B equals the change in kinetic energy from A to B. Regarding the trig, draw a sketch.
  4. Jan 27, 2007 #3
    ok then the loss in KE=1000J
    but thats not the right answer, theres also resistance forces how do i take that into account and what do you mean by internal and external forces.

    Last edited: Jan 27, 2007
  5. Jan 27, 2007 #4


    User Avatar
    Homework Helper

    work done by pulling force + work done by resistance force = change in kinetic energy. => (iii) work done by pulling force = change in kinetic energy - work done by resistance force. This should do.
  6. Jan 27, 2007 #5
    work done by pulling force = 1000 -(7.5x200)=-500 J

    i though it was

    work done by pulling force= gain in ke + gain in ke + work done by resisting force

    but if its loss in ke then we minus it
  7. Jan 27, 2007 #6
    just bump this up.
  8. Jan 27, 2007 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Let's take one question at a time:

    How would you calculate the answer to i)?
  9. Jan 28, 2007 #8
    green pandas

    loss in KE=1/2m(v^2-u^2)=

    Gain in PE=mgh=(50)(10)(15)=7500J

    so has the car goes up the straight hill it gains 7500J in PE but looses 1000J KE.
    Because thers a resistance force of 7.5 N, the work done by it=fs=7.5(200)=1500J

    So work done py pulling force=(7500-1000)+(1500)=8000J
  10. Jan 28, 2007 #9
    thats wrong

    work done by pulling force - work done by resistance=change in kinetic energy
  11. Jan 28, 2007 #10


    User Avatar
    Homework Helper

    If I wrote '+', that doesn't mean that the work done by resistance doesn't have a '-' to reveal behind its name. :wink:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?