# Initial Velocity Question

## Homework Statement

A 20.00kg curling stone is released at the hog line and moves 28.35m [W] to sit on the button and score a point. If the coefficient of kinetic friction between the stone and the ice is 0.0020000, what was the initial speed of the stone ?

So we know:
m=20.00kg
d=28.35m[W]
Fk=0.0020000
Vi=?

## The Attempt at a Solution

I'm stuck on what I have to find first to end up finding the velocity. Once I know what I need to find first I can figure it on from there. I am not asking for any answers, just what I have to find first. Any help is appreciated.

What principle relates work to energy?

Force of friction ?

Or are you studying distance, acceleration equations?

Distance and acceleration equations. We haven't learned anything with energy yet.

This problem can be solved by a velocity, distance, acceleration approach or by energy considerations. Which is it?

Ok, our posts crossed. Do you know of an equation that relates velocity to distance and acceleration?

Ok, our posts crossed. Do you know of an equation that relates velocity to distance and acceleration?

The only equation I know that has velocity, acceleration and distance is :
Δd=v1Δt + 1/2 aΔt^2

You have d = 1/2 * a * t^2. You also have V = a * t. Using both equations, eliminate the time variable.

t=v/a
2d/a=t^2

Plug the expression you have for t into the distance equation. That gives you a relationship between velocity, distance, and acceleration. You can get the acceleration from Newton's Law considering friction and mass and gravity.

I have to be signing off now. If you are still having difficulty, hang in there. There are plenty of competent helpers on this forum who will be glad to help out.

Plug the expression you have for t into the distance equation. That gives you a relationship between velocity, distance, and acceleration. You can get the acceleration from Newton's Law considering friction and mass and gravity.

I have to be signing off now. If you are still having difficulty, hang in there. There are plenty of competent helpers on this forum who will be glad to help out.

Ahhh, okay. Thank you for your help !