# Finding Force Applied on a ball from another object

• Drake M
In summary, the conversation discusses the calculation of the applied force on a pool ball, taking into account factors such as force of friction, distance traveled, and mass of the cue and ball. There is confusion about which equations to use and whether to consider the friction as static or kinetic. The main goal is to determine the applied force and its magnitude. The conversation also mentions a concern about the high coefficient of friction in relation to a pool table.
Drake M

## 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

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

First, 2cm is one hell of a contact distance for a pool shot.

Your calculation for the applied force seems to assume that the stroke causes the cue to come to rest, thereby using all of its KE. Would that be right? There was no tendency for the player to arrest the cue?

Using W=Fd to find F only really works if the force is constant. If it is not constant then it will not even give you the average force. But that's a minor point here.

## 1. How do you calculate the force applied on a ball from another object?

The force applied on a ball from another object can be calculated using the formula F = m x a, where F is the force, m is the mass of the ball, and a is the acceleration caused by the other object.

## 2. What is the difference between force and applied force?

Force refers to a push or pull that can cause an object to move, while applied force specifically refers to the force applied to an object by another object.

## 3. Can the force applied on a ball from another object change?

Yes, the force applied on a ball from another object can change if the mass or acceleration of either object changes. For example, if the mass of the ball increases, the force applied on it by another object will also increase.

## 4. How does the distance between two objects affect the force applied on a ball?

The force applied on a ball from another object is inversely proportional to the square of the distance between the two objects. This means that as the distance between the two objects increases, the force applied on the ball decreases.

## 5. What is the unit of measurement for force?

The unit of measurement for force is Newton (N) in the International System of Units (SI). In the British system, the unit of force is pound (lb).

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