Question

Please help! I don't know how to solve this question!
A shuffle board disk is accelerated to a speed of 5.5 m/s and released. If the coefficient of kinetic friction between the disk and concrete is 0.34, how far does the disk go before it comes to a stop?

Answer 1

1) Solve for the acceleration
\[a=\mu \times g\]
2) Use the constant acceleration formula to solve for the distance
\[v^2=u^2+2ax\]
- v is the final velocity
- u is the initial velocity
- a is the acceleration(found previously)
- x is the distance

Answer 2

I haven't learned that formula yet

Answer 3

The first one?

Answer 4

Both

Answer 5

What sort of formulas have been using previously to this question?

Answer 6

I use Fg=mg Ff= uFn a= (Fa-Ff)/m F=ma for all my friction problems

Answer 7

The first formula I showed above actually comes from the formula 'Ff=uFn'. So you can use it in your question.
\[F _{f}=\mu \times F _{n}\]
- The frictional force(Ff) for an object moving linearly can be found by multiplying its mass with acceleration.

- The Normal force(Fn) is equal to the Weight force which is the object's mass multiplied by gravity.

Once we substitute this information into our formula, we get:
\[ma = \mu \times mg\]
Notice that m is common on both sides of the equation and we can cancel it out. So we are left with:
\[a=\mu \times g\]
and thats where that formula comes from.

Answer 8

O ok, but how do I find the distance?

Answer 9

For distance, use the second formula I showed you.
\[v^2=u^2+2ax\]
- v is the final velocity. It says in the question that the disk comes to a stop. So the final velocity(v) is equal to 0
- u is the initial velocity given to you in the question
- a is the acceleration which comes from the formula a=ug
- x is the distance that you are trying to find. Rearrange the equation to solve for x.

Answer 10

Thank you for helping

Answer 11

My pleasure!

Answer 12

This

Answer 13

I need to figure this out help me out

Answer 14

i give you a meadel