top top a Essentially frictionless, horizontal ice rink, a skater relocating in ~ 6.0 ms-1 encounters a turbulent patch that reduces her rate by 40% as a result of a friction pressure the is 26% the her weight.

You are watching: Use the work-energy theorem to find the length of this rough patch.

so what ns tried to perform ins together follows:

WD = KE(Final) - KE (Initial) = 0.5m(3.6)^2 - 0.5m(6)^2 = 6.48m - 18m = -11.52mfor this reason that'ns the work-related Da by ns friction force (Let's contact ins Ff), which is same to -0.26m*d.

for this reason then ns get,

-11.52m = -.26m*street distance = -11.52/-.26 = 44 metres.Now, ns know wcap I've da is wrong, yet I'm no certain wright here mine mistake it is. There'ns a similar difficulty in ns Publication the ns additionally do the efforts because that practice (It just readjusted the number around a little) but i can't gain the ideal price for the either. In case it helps, the one in the Publication supplies 3 ms-1 for Initial V, 1.65 ms-1 for final V and a friction force equal to 25% the her weight, and the answer ins 1.3 m.


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level 1
· 9y · edited 9y
Engineering, Biology, Physics, Statistics
well ins seems you lefns out a large component that ns question, yet ns will still try and also help. First, you should gain her devices straight, you are making use of fifty percent that them, when omittinns the rest. The speed to be in systems m/s. Also, your KE formula is wrong, KE is .5(m)(v)2, friend forgained come encompass ns mass.

WD = KE(Final) - KE (Initial)

= 0.5(m)(3.6)^2 - 0.5(m)(6)^2 = 6.48m - 18m = -11.52mNow, the frictional pressure is equal come 26% the ns skater'ns weight, not your mass. Thins amounts to .26(m)(g) = 2.5506m

Keep in mind that g = 9.81 m/s2, and also Keep in mind the the work-related done by the frictional force ins negati have Because it actns versus ns skater's movement.

native there, i thoctopus you have the right to number ins out. Just make certain to match units and you'ltogether it is in fine. Lens me recognize if girlfriend require any more help.


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levetogether 2
Op · 9y

got it, thanks a bunch


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levetogether 2
Ons · 9y

the was the whole question, word because that word.

ns have actually m contained in ns formula, however I'm not offered a mass, for this reason i can'ns pluns it in, for this reason I've just left ins together 'm'. I thsquid the ns m in ns KE will cancel via ns m in the frictional force equation, however obviously the means i've da ins hasn'ns functioned out.


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proceed thins thread 


level 2
Op · 9y
I'ltogether try this, thanks


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level 1
· 9y

the mistake is the frictional pressure is no -0.26md, it's -0.26mg, Since it wtogether 26% that the weight. Thus ns friction job-related da ins -0.26mgd. Therefore you simply have to division her price by ns = 9.8 m/s2 and friend acquire a reasonmay be prize that 4.52 m.

See more: The Group Or Resource Is Not In The Correct State, To Perform The Requested Operation


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