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!

Law of conservation of energy

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.

    In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.

    In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object's kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation

    K_i + U_1 + W_other = K_2 + U_2 ,

    where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.


    What is the speed of the object at the height of ?
    Express your answer in terms of and . Use three significant figures in the numeric coefficient.

    2. Relevant equations

    K_i + U_1 + W_other = K_2 + U_2

    3. The attempt at a solution

    (1/2) mv^2 + 0 + 0 = (1/2) mv^2 + mg (v^2 / 4g)

    so when I solve for v it = 0

    mv^2 = 2(0)

    v = 0

    What did I do wrong?
     
  2. jcsd
  3. Feb 28, 2008 #2
    So you're assuming it launches from the ground, so Ki=1/2*mVi^2

    then at some other point it will have slowed down of course, so Kf=1/2*m*Vf^2, and the potential energy will be U=mgh

    You tried to solve for h as a function of its velocity(which at that point will be the same Vf) as in the kinetic energy equation but I don't believe you did it right
     
  4. Feb 28, 2008 #3

    Astronuc

    User Avatar

    Staff: Mentor

    You need to find that appropriate expression for the gravitational potential energy, U_1 and U_2, as functions of r, where r is the radius from the center of the mass responsible for the gravitational field.

    http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#gpt
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Law of conservation of energy
Loading...