# Forces AND Friction AND Motion O MY!

1. Oct 19, 2009

### kza62

1. The problem statement, all variables and given/known data

A cart with a mass of 285 kg is pushed horizontally with a force of 1400 N. The acceleration of gravity is 9.8 m/s/s. The coefficient of friction (u) with the crate and the surface is 0.29. Calculate the acceleration of the crate.

Sooo given:

Fa: 1400 N
m: 285 kg
Fg: -9.8
u: 0.29
Ff: ?
Fn: ?
a: ?

2. Relevant equations

Fnet = ma
Ff = u * Fn

3. The attempt at a solution

Ok....
so if Fg is -9.8 then Fn is 9.8?
and then Ff = -(0.29)(9.8) which is -2.842?
so would Fnet = 1397.158 N?
and then do I use F = ma to find a?
so a = F/m which i got 4.9023
but the answer in the back of my text book is 2.07028!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 20, 2009

### tiny-tim

Hi kza62!
no no no no noooo

-9.8 is the acceleration, isn't it?

3. Oct 25, 2009

### kza62

not really dude, its moving horizontally, its not falling

4. Oct 26, 2009

### tiny-tim

Yes, but I meant that when you wrote Fg = -9.8, I assumed that your F meant a force, so that would be -9.8m

5. Oct 26, 2009

### kza62

ooooo okokokok but how do u get Fn?? thats the only thing i need to plug in everything else :uhh:

6. Oct 27, 2009

### tiny-tim

(just got up :zzz: …)

The ground is horizontal, so Fn (the normal force) is the same as Fg, and then Ffriction = µFn.

You know that, don't you?

7. Nov 1, 2009

### kza62

so Ff = 0.29 x 9.8 = 2.842
and then Fnet = 1400 - 2.842 = 1397.158
a = 1397.158/285 = 4.902308... ????

what did i do wrong now???!!!

8. Nov 1, 2009

### tiny-tim

I've told you … you have to use 9.8m.