# Newtons Laws

1. Nov 20, 2009

### tigerwoods99

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

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.18 and the push imparts an initial speed of 4.1 m/s?

2. Relevant equations

MG of the object = mass * 9.8
Friction = mg * 0.18

3. The attempt at a solution

//

ANY HELP WOULD BE MUCH APPRECIATED

2. Nov 20, 2009

### tiny-tim

Hi tigerwoods99!

Use the work-energy theorem …

work done = change in energy

3. Nov 20, 2009

### tigerwoods99

sorry, im not familiar with this theorem

4. Nov 20, 2009

### tiny-tim

oops!

In that case, find the acceleration from µ = 0.18, and then use one of the standard constant acceleration equations, with vi = 4.1 and vf = 0.

5. Nov 20, 2009

### tigerwoods99

That would make sense. I know the acceleration has to be negative because it comes to a stop, the acceleration is -> (direction) and the friction is <- (direction)

Are these the formulas I could use to find the acceleration using mu? I have a feeling i have to know the weight of the object though to find the mg and fn
a = Fnet/m
Ffriction = Fnormal * mu

Vi: 4.1 m/s
Vf: 0 m/s
D:
A:
T:

6. Nov 20, 2009

### ideasrule

If a=Fnet/m, Fnet=Fnormal*mu, and Fnormal=mg, then a=?

7. Nov 20, 2009

### tigerwoods99

yes, i understand the formulas but how do i get the values for the different forces, if the mass of the object is unknown?

8. Nov 20, 2009

### ideasrule

Just assume the mass is m and try it. You'll find that m cancels out.

9. Nov 20, 2009

### tigerwoods99

A = Fnet/m
A = (Fnormal *mu)/m

A = (Fnormal *u) ?
So how would I find the FN

10. Nov 20, 2009

### ideasrule

Fn exactly balances gravity, or else the object would accelerate in the y direction. So Fn=mg.

BTW:

A = (Fnormal *mu)/m
A = (Fnormal *u) ?

11. Nov 20, 2009

### tigerwoods99

mu = mg/9.8 * u

12. Nov 20, 2009

### ideasrule

"mu" is a single constant, representing the coefficient of friction. It is not m*u, so mu/m isn't equal to u (which is meaningless).

13. Nov 20, 2009

### tigerwoods99

thats what i thought, but wasn't sure becuase i am used to seeing it as just u

14. Nov 20, 2009

### tigerwoods99

a = fnet/m
a = (fnormal * mu)/ m
a = (mg * mu)/m
a = gravity * mu

but becuase the object is moving vertically there is no acceleration

15. Nov 21, 2009

### ideasrule

Yes, that's right

16. Nov 21, 2009

### tiny-tim

Hi tigerwoods99!

(just got up :zzz: …)
(oh, if only everybody had a Mac instead of a PC, with a sensible keyboard! )

have a mu … µ
(try using the X2 tag just above the Reply box )

Are you confusing the vertical and horizontal accelerations?

Vertically, a = 0, and Fnet = N - mg, so N = mg.

Horizontally, Fnet = µmg, so ma = µmg.

17. Nov 21, 2009

### tigerwoods99

thanks i got it!