# Forces Problem (find acceleration of bottle and how far it travels from rest.)

## Homework Statement

David pushes on a .50 kg bottle of water with a force of 3.5 N (R). The coefficient of friction between the bottle and the desk is .20. Using the dynamics worksheet format below, find the acceleration of the bottle and how far it travels in 3.0 s if it starts from rest.

## Homework Equations

Not sure which ones to use:
-> Fnet=Fa-Ff
-> Fa = 3.5 N (E)
-> Ff= coefficient of friction*Fg
-> Fg= mg

Forces acting on bottle: Fg, Fn, Ff, Fa

## The Attempt at a Solution

Fa=3.5N
coefficient of friction=.20
a=?
t=3.0s
V2=?
V1= 0m/s^2

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PhanthomJay
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## Homework Statement

David pushes on a .50 kg bottle of water with a force of 3.5 N (R). The coefficient of friction between the bottle and the desk is .20. Using the dynamics worksheet format below, find the acceleration of the bottle and how far it travels in 3.0 s if it starts from rest.

## Homework Equations

Not sure which ones to use:
-> Fnet=Fa-Ff
-> Fa = 3.5 N (E)
-> Ff= coefficient of friction*Fg
-> Fg= mg

Forces acting on bottle: Fg, Fn, Ff, Fa

## The Attempt at a Solution

Fa=3.5N
coefficient of friction=.20
a=?
t=3.0s
V2=?
V1= 0m/s^2
Don't forget Newton's 2nd law in your relevant equations (F_net =ma). Then you can solve for the acceleration. Also, how about then using the kinematic equation of motion that relates distance with time and acceleration?

Using Ff = uFN, find force of friction. So Ff = (0.20)(0.50)(9.8) = 0.98 N
Find the net force by subtracting 0.98 N from 3.5 N = 2.52 N.
Then using F=ma, find a. 2.52/0.50 = 5.04 m/s^2

Using d= vit + 0.5 at^2, you can find d.

so d = (0)(3) + (0.5)(5.04)(3^2) = 22.68 m.

This this right

PhanthomJay Looks real good!