Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two Energy Problems

  1. Oct 22, 2005 #1
    Conservation of Mechanical Energy
    (1) A vertical spring (ignore it's mass), whose spring stiffness constant is 950 N/m, is attached to a table and is compressed down 0.150m. (a) What upaward speed can it give to a 0.30kg ball when released? (b) How high above it's orginal position (Spring compressed) will the ball fly?
    (a) Okay this is what I did for part a, and my answer was close but not close enough...
    1/2kx^2 = 1/2mv^2
    .5*950N/m*.150m^2 = .5*.30kg*v^2
    v = 8.4 m/s (but the book gave the answer of 8.3 m/s) did I do anything incorrect?
    (b) 1/2Kx^2=mgh
    .5*950*.150^2 = .3*9.8*h
    h = 3.64 m <~~~ I got that one correct.
    Law of Conservation of Energy
    (2) A ski starts from rest and slides down a 22 degree incline 75 m long.
    (a) If the coeffiecent of friction is .090, what is the ski's speed at the base of the incline?
    (b) If the snow is level at the foot of the incline and has the same coeffiecent of friction, how far will the ski travel along the level? Use energy methods.
    I need some to help me with problem 2. I have drew free bodies already... can't figure out how to attempt it.
  2. jcsd
  3. Oct 22, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You forgot to include a very important type of potential energy in your equation for energy conservation here..

    As for 2, what work must be done against the friction?
    Last edited: Oct 22, 2005
  4. Oct 22, 2005 #3
    A Very important type of PE? Huh? Thereis a missing variable?
  5. Oct 22, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, you haven't included the change in GRAVITATIONAL potential energy from when the spring is compressed and when it is not.
  6. Oct 23, 2005 #5
    So maybe...

    mgh + 1/2kx^2 = 1/2mv^2
    (.30kg*9.8*.150m) + (.5*950N/m*.150m^2) = .5*.30kg*v^2
    11.1285 = .15 v^2
    sqr of 74.19 = 8.61 <~~~ that isn't right...
    Last edited: Oct 23, 2005
  7. Oct 23, 2005 #6
    I think I may have got it... I believe I added the PE incorrectly. The equation should read:

    1/2kx^2 = 1/2mv^2 + mgy

    (.5*950N/m*.150m^2) = .5*.30kg*v^2 + (.30kg*9.8*.150m)
    10.2465 = .15v^2
    sqr of 68.31 = 8.26 = 8.3 m/s

    so, that is the correct way?
  8. Oct 24, 2005 #7
    (b) 1/2Kx^2=mgh
    .5*950*.150^2 = .3*9.8*h
    h = 3.64 m

    that is correct also?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook