Question

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

Answer 1

sorry, saw you were off line so thought you were gone for the evening...

Answer 2

let me look it over..

Answer 3

sure, so...
work done by the rough spot (friction) = 1/2*11*( 1.5^2 -2.8^2)

Answer 4

length of the rough spot is from Work = F*d = mu*mg*d

Answer 5

now the last part should be clearer... it's a little tricky but not too tough...

Answer 6

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

Answer 7

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

Answer 8

work energy theorem problems? you should be able to do them now... I'll answer any specific questions you have about particular details...

Answer 9

now i have conservative forces and potential energy problem