- #1

Drake M

- 19

- 0

## Homework Statement

Another question regarding pool. I need to know how much force applied is given from the cue to the ball as well as force of friction

OF CUE:

a=56.55m/s2

vi=0m/s

vf=3.28m/s

d=.0953m

m=.55kg

BALL:

m=.17kg

*not sure whether to use distance traveled during contact or of the ball after it was contacted as well as if to use vf of the cue as vi of the ball

## Homework Equations

vf*2=Vi*2+2ad

W=Fd

Fnet=Fa+Ff

Co-efficient F= Ff x Fn

## The Attempt at a Solution

For FNET of the ball i found acceleration using .02m as distance traveled during contact as well as vi=0m/s and vf=3.28m/s. Using vf*2=Vi*2+2ad I got an a of 268.75m/s2

Then using fnet=ma, I fot 45.65N as fnet of the ball. I also did the Ek=(1/2)mv*2 then w=fd to find fnet but results were the same.

Where I'm completely stuck is how to find the applied force on the ball.

I went and found fnet of the cue using Ek equation (m=.55kg, vf=3.28m/s) then used w=fd(d=.0953m) and got 30.95N. I felt good but then used FNET=fa+ff to get force of friction and it came out as 14.7N which wouldn't be weird if it weren't for the fact that it is greater than the normal force so the coefficient of friction is larger than 1. I know that its possible but having that high of a coefficient doesn't make sense for a pool table.

Main goal is to find out what the applied force was and how much it is. ALSO, FORGOT TO MENTION IF THE FRICTION IS STATIC OR KINETIC. Thanks in advance to everyone