Energy and Work Energy and Work Physics Help Forum Dec 9th 2016, 01:15 AM #1 Junior Member   Join Date: Dec 2016 Posts: 2 Is this Power/Work problem Possible I am currently in AP Physics C. I was given a homework problem that I don't think is possible with the given info. A 20 kg rock slides on a rough horizontal surface at 8 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average thermal power is produced as the rock stops? (Answer: 157 WATTS) (first get the work done!) Here I am given mass, velocity, force of friction( (20 * 9.8) * .2 = 39.2), AND THAT's IT. I tried getting work first, but I don't have a distance. I suppose I could solve for acceleration since friction is the only force in the X direction, which is -1.96, but even then, how would I get a distance from that. If I had a period of time this took place over I could find distance, but I don't. Could somebody explain this? Thanks   Dec 9th 2016, 07:46 AM #2 Senior Member   Join Date: Jun 2010 Location: NC Posts: 417 Similar Problem... Hi LPB, Is it (v2)(v2) - (v1)(v1) = 2 ax? Does that get "x." Interesting problem. I wonder if it is the average "frictional" power that happens? Here's a friction loss problem. Not your problem but something to look at. Crate Pushed Uphill | THERMO Spoken Here! Good Luck, JP   Dec 9th 2016, 08:10 AM   #3
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 Originally Posted by THERMO Spoken Here Hi LPB, Is it (v2)(v2) - (v1)(v1) = 2 ax? Does that get "x." Interesting problem. I wonder if it is the average "frictional" power that happens? Here's a friction loss problem. Not your problem but something to look at. Crate Pushed Uphill | THERMO Spoken Here! Good Luck, JP
I don't think that's it, but seeing that formula made me remember what to do. It's a kinematics equation. Vfinal (0) = vinitial(8) + acceleration(1.96)*time. From there I can solve for time and get the answer, thanks.   Dec 9th 2016, 08:59 AM #4 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,343 Right, the initial KE is all converted to heat, so the average power is Power =$\displaystyle \frac {\Delta KE} t = \frac {\frac 1 2 mv^2 }{t}$ The time for the rock to stop comes from $\displaystyle \Delta v = at$. You just need to determine 'a' from $\displaystyle F_r = \mu mg = ma$.  Tags friction, naka, power, power or work, problem, work Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Alexis87 Advanced Mechanics 4 Apr 3rd 2013 08:20 AM unwillingengineer Energy and Work 3 May 12th 2010 10:47 PM strgrl Energy and Work 9 Sep 29th 2009 12:53 AM angelica Kinematics and Dynamics 9 May 20th 2009 10:22 PM angelica Kinematics and Dynamics 3 May 14th 2009 12:33 AM 