Respuesta :
Answer:
5.53 m
Explanation:
[tex]v_{o}[/tex] = initial speed of shuffleboard disk = [tex]5.8 ms^{-1}[/tex]
[tex]v_{f}[/tex] = final speed of shuffleboard disk = [tex]0 ms^{-1}[/tex]
[tex]\mu[/tex] = Coefficient of kinetic friction between the disk and concrete court = 0.31
acceleration due to friction is given as
[tex]a = - \mu g\\a = - (0.31) (9.8)\\a = - 3.04 ms^{-2}[/tex]
[tex]d[/tex] = distance traveled by disk before it stops
using the kinematics equation that fits the above the data, we have
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 5.8^{2} + 2 (- 3.04) d\\d = 5.53 m[/tex]
Answer:
5.54 meters
Explanation:
First you need to derive the equation to find acceleartion.
STEP BY STEP:
Fnet=Fn-mg-Fk
Fnet=ma so we can say:
ma=Fn-mg-Fk
(Fn=mg) and (Fk= μFn)
ma=mg-mg-(μmg)
divide by m on all sides and get
a=g-g-(μg)
since we know that Fn=mg we can replace Fn by mg^
a=g-g-(μg)
a=g(-μ+1-1)
a=g(-μ) or a=-μg
so you have your equation for acceleartion find your a and put it into the equation and get -3.038
Vf^2=Vi^2+2ad
0^2=5.8^2+2(-3.038)d
Solving for d we get 5.54 meters
