If a #4 kg# object moving at #1 m/s# slows down to a halt after moving #90 m#, what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2016-02-16T07:31:34

#5 * 10 ^-4#

#2as = v^2 - u^2#

#a =( Deltav^2)/(2s) = 1 / 180#

#f = mu N#

#N = mg = 40 N#

#D_f = mu N/m = mu g#

#mu g = 1/180#

#mu = 1/1800 = 0.00055555555 approx 5 * 10 ^-4#