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!

Energy and Momentum

  1. Jul 5, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider a frictionless roller coaster.
    Point A=95m
    Point B=65m
    Point C=65m
    Point D= 25m
    Point E= Is 7m from the end of the course where the height is 0

    If a 1200kg car starts at rest from point A, calculate:

    a)The total energy of the system
    b)The speed of the car at point B
    c)The force that must be applied to bring it to a stop at point E
    d)The work done to bring it to a stop at point E

    The attempt at a solution

    a)
    The total energy in a system is the sum of all its kinetic energy and potential energy
    Et=total energy

    Et=½mv^2 + mgh

    =½(1200kg)(0)^2 + (1200kg)(9.8m/s)(95)
    =0 + 1117200
    =1.1 x 10^6 J

    b)
    m=1200kg
    d=(95m-65m)=30m
    g=9.8m/s

    vf^2=vi^2 + 2aΔd

    =0+2(9.8m/s)(30m)
    =√588
    =24 m/s

    Ok, now question c) and d) is where I get stuck.

    I think I need to use this equation W=FcosΘΔd for part d)...but I'm not sure.

    Any help for question c) and d) would be greatly appreciated. Thank you :)
     
  2. jcsd
  3. Jul 5, 2015 #2
    Forget momentum. You got lucky on B. You should really use conservation of energy rather than a kinematic equation.

    C and D, you need to use the work-energy theorem.
     
  4. Jul 5, 2015 #3
    Ok so is that W = KEf - KEi? Where KE = ½mv^2.

    So Would it be W=½mv^2f - ½mv^2i

    W=½(1200kg)(0)^2 - ½(1200kg)(43m/s)^2

    =0 - 1109400
    = -1109400 J
    = -1.1 x 10^6 J

    Is this closer for d)?
     
  5. Jul 5, 2015 #4
    That's the right idea.
     
  6. Jul 5, 2015 #5
    Thanks,
    Im still not sure on how to solve for force though. This is pretty new to me.

    for part d) since -1.1 x 10^6 J of work was done to bring the car to a stop at point E, would 1.1 x 10^6 J of force have to be applied to bring the car to a stop for part c)?
     
  7. Jul 5, 2015 #6
    No, Energy is measured in Joules (or Newton Meters). Force is measured in Newtons.

    You have used the Work-Energy theorem to compute the work.

    Now remember how to relate work, distance, and force to compute the force.
     
  8. Jul 5, 2015 #7
    No, Energy is measured in Joules (or Newton Meters). Force is measured in Newtons.

    You have used the Work-Energy theorem to compute the work.

    Now remember how to relate work, distance, and force to compute the force.
     
  9. Jul 5, 2015 #8
    Ok,

    W=F•Δd
    ∴F=(W) / Δd

    F= (1109400 J) / (7m)

    =158485.71 N
    =1.6 x 10^5 N

    better?
     
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: Energy and Momentum
  1. Momentum, Energy (Replies: 7)

  2. Energy momentum (Replies: 1)

  3. Momentum + Energy (Replies: 8)

  4. Energy and momentum (Replies: 0)

  5. Energy and momentum. (Replies: 7)

Loading...