Physics Help Forum Conservation of energy problem

 Energy and Work Energy and Work Physics Help Forum

 Oct 17th 2017, 09:04 AM #1 Junior Member   Join Date: Oct 2017 Posts: 8 Conservation of energy problem If the speed of a 62 kg skier at P is 1.5 m/s, and the conversion from potential energy to kinetic energy as she skies down the hill is 65% efficient, then her speed at point Q will be _ m/s and _ kJ of energy will have been lost due to friction. I found the speed and it was 11.57 m/s. However, I don't get the same answer as the answer key says for the kJ of energy lost due to friction. The answer was 2.6 kJ, but I keep getting 3.6 kJ. What I did was I did epi+eki= ekf + epf + heat. Isolated heat to find the energy lost due to friction. I used the speed, 11.57m/s for the ekf. I still keep getting 3.6kJ and not 2.6 at all.
 Oct 17th 2017, 10:31 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,299 I get 2.6 KJ, but I don't agree that the final velocity is 11.57 m/s. Please show us how you arrived at that. I get a final velocity of 12.45 m/s, from: $\displaystyle \Delta PE \times 0.65 = \Delta KE$ $\displaystyle m g (15m -3m) (0.65) = \frac 1 2 m (v_2^2 - v_1^2)$
 Oct 17th 2017, 10:46 AM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 987 change of PE = g x mass x change of elevation = 9.81*62*(15 -3) Joules = 7298.64J 35% of this is lost 7298.64 * 0.35 = 2.554J or 2.6 kJ 65% of the lost PE ends up as increased KE. The skier starts off 0.5 *62 * 1.5 * 1.5 J of KE = 69.75J and gains 7298.64*0.65 J = 4744.12J of KE Thus she has a total of 4813.87J of KE. This corresponds to a velocity of sqrt (4813.87/31) = 12.46 m/sc Edit I see I cross posted with Chip.
 Oct 17th 2017, 11:04 AM #4 Junior Member   Join Date: Oct 2017 Posts: 8 I did 0.65= ((.5*62*vf^2) + (62*9.8*3))/((.5*62*1.5^2) +(62*9.8*15)). Isolated vf to get 11.57.
Oct 17th 2017, 11:38 AM   #5
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 987
 Originally Posted by Latinized I did 0.65= ((.5*62*vf^2) + (62*9.8*3))/((.5*62*1.5^2) +(62*9.8*15)). Isolated vf to get 11.57.
I can't tell from that reply whether you have realised your mistake or not?

Oct 17th 2017, 12:07 PM   #6
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 987
 65% of the lost PE ends up as increased KE.
I see I misphrased this line, which should have read

65% of the change in PE ends up as increased KE.

sorry.

 Oct 17th 2017, 12:18 PM #7 Junior Member   Join Date: Oct 2017 Posts: 8 I saw that both you did it a different way, but I dont understand why you'd do it that way.
Oct 17th 2017, 12:45 PM   #8
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 987
 Originally Posted by Latinized I saw that both you did it a different way, but I dont understand why you'd do it that way.
Actually both of us did it basically the same way.

Have you understood where you went wrong yet?

Hint this is not correct reasoning

 I did 0.65= ((.5*62*vf^2) + (62*9.8*3))/((.5*62*1.5^2) +(62*9.8*15)). Isolated vf to get 11.57.

 Oct 17th 2017, 03:32 PM #9 Senior Member   Join Date: Jun 2010 Location: NC Posts: 398 Past Problem Hello Latinized, Your problem is similar to one submitted to PHF some months past. I solved it (as best I could). I reference it for your information. Boy in a Wheelchair | THERMO Spoken Here! Good Luck... JP topsquark likes this.

 Tags conservation, energy, problem

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Charles Energy and Work 4 Dec 15th 2015 08:19 AM Momentous Energy and Work 6 Jul 8th 2013 12:01 PM Nibbz Energy and Work 2 Nov 20th 2009 09:37 AM alexito01 Advanced Mechanics 1 Dec 15th 2008 04:40 PM shogunhd Advanced Mechanics 1 Dec 10th 2008 06:10 PM