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)
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?
4 answers
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
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.
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
1 answer