In a game of shuffleboard, a disk with an initial speed of 3.2 m/s travels 6.0 m before coming to rest. (a) What was the magnitude of the average acceleration of the disk? (b) What was the coefficient of kinetic friction acting on the disk?
2006-10-03 12:55:26 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
i have no idea..i study philosophy...
2006-10-03 12:56:50 · answer #1 · answered by Anonymous · 0⤊ 1⤋
For constant acceleration
Îx = [v(f)^2-v(i)^2]/2a
or
a = [v(f)^2-v(i)^2]/2Îx
so a = [0^2-(3.2m/s)^2]/2*6.0m
a ~ - .853m/s^2
So magnitude or |a| ~ .853m/s^2
μ = F/N = m|a|/mg = |a|/g = .853m/s^2 / 9.8m/s^2
μ ~ .087
2006-10-03 20:10:16 · answer #2 · answered by Andy S 6 · 0⤊ 0⤋
distance=6=3.2t+at^2/2
a=-3.2/t
substitute
6=3.2(-3.2/a)+a/2*(-3.2/a)^2=3.2^2(-1/2)/a
a=-3.2^2/(2*6)=-.853
part a) magnitude of a=.853 m/s^2
coeficient of friction:
gravity" 9.8 m/s^2
.853/9.8=.087
2006-10-03 20:32:55 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋