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A mass m = 11.0 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5008.0 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.46. The mass leaves the spring at a speed v = 2.8 m/s. 3) The mass is measured to leave the rough spot with a final speed vf = 1.5 m/s. How much work is done by friction as the mass crosses the rough spot? What is the length of the rough spot? In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed fro

Mathematics
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sorry, saw you were off line so thought you were gone for the evening...
let me look it over..
sure, so... work done by the rough spot (friction) = 1/2*11*( 1.5^2 -2.8^2)

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Other answers:

length of the rough spot is from Work = F*d = mu*mg*d
now the last part should be clearer... it's a little tricky but not too tough...
find W= mu*mg*d/2 that's the KE the mass had when it encountered the patch (and also the KE when it left the spring) KE=PEspring mu*mg*d/2 = 1/2 *k*x^2
thank you soo much ur a life savor litterally..i have 2 more questions i cant get .if you wouldnt mind helping me on those id appreciate it soo much
work energy theorem problems? you should be able to do them now... I'll answer any specific questions you have about particular details...
now i have conservative forces and potential energy problem

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