Asked by Farah
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?
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?
Answers
Answered by
bobpursley
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)
Initial energy-frictionwork=final energy
friction work= initial energy-finalenergy
= 1/2m(14.3^2)+mg*2 - 1/2 m (15.6^2)
Answered by
bark
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
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
Answered by
bobpursley
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.
Answered by
bark
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
i forgot the units, but sure it is joule