Physics - conservative and non conservative (check

If a sled reaches the base of the hill with a speed of 15.6m/s, how much work was done by the snow on the sled between points x and y, where x is 3m above the ground with a velocity of 14.30m/s and y is toughing the ground. the sled and the rider are 70kg

Ans: i tried the formula work done = Ek+Ep, where Ek =(0.5)(70)(15.60)^2 and Ep= (70)(9.81)(2) i get 7157j which is incorrect, where am i going wrong?

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asked by Farah
  1. Hmmm. It has an initial PE, and an initial KE. Then at the bottom it has only KE. So you want to find the work done on friction.

    Initial energy-frictionwork=final energy

    friction work= initial energy-finalenergy
    = 1/2m(14.3^2)+mg*2 - 1/2 m (15.6^2)

  2. Epot=mgh = (70)(9.81)(3)=2060,1
    Ek1 =(0.5)(70)(14.30)^2= 7157,15
    Ek2 =(0.5)(70)(15.60)^2=8517,6

    total energy= (Epot+Ek1)- Ek2
    =9217,25-8517,6
    =699,65

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    posted by bark
  3. I object to your use of "total energy". Your formula expression is the same as mine, but I found the work done on friction. That is not, nor even close to, total energy. So I am not certain what you mean. You found friction work. I also object to the number of significant digits, carrying these calculations to 5 or 6 places is lubricious. You also should use units, the final answer is in Joules (J). I did not check your calculator work.

  4. i agree with you but is the result correct, my english is'nt ok i know

    i forgot the units, but sure it is joule

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    posted by bark

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