Question

A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop?

Answer 1

5.54 meters

Step-by-step 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

Answer 2

5.53 m

Step-by-step explanation:

`v_(o)` = initial speed of shuffleboard disk =
`5.8 ms^(-1)`

`v_(f)` = final speed of shuffleboard disk =
`0 ms^(-1)`

`\mu` = Coefficient of kinetic friction between the disk and concrete court = 0.31

acceleration due to friction is given as

`a = - \mu g\\a = - (0.31) (9.8)\\a = - 3.04 ms^(-2)`

`d` = distance traveled by disk before it stops

using the kinematics equation that fits the above the data, we have

`v_(f)^(2) = v_(o)^(2) + 2 a d\\0^(2) = 5.8^(2) + 2 (- 3.04) d\\d = 5.53 m`