Sliding Box: Determining Distance Traveled with Newton's Laws

[tigerwoods99]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

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ANY HELP WOULD BE MUCH APPRECIATED

 

[tiny-tim]

Hi tigerwoods99!

Use the work-energy theorem …

work done = change in energy ​

 

[tigerwoods99]

sorry, I am not familiar with this theorem

 

[tiny-tim]

sorry, I am not familiar with this theorem

oops!

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

 

[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:

 

[ideasrule]

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

 

[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?

 

[ideasrule]

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

 

[tigerwoods99]

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

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

 

[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) ?

Think about that. What's mu?

 

[tigerwoods99]

mu = mg/9.8 * u

 

[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).

 

[tigerwoods99]

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

 

[tigerwoods99]

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

but because the object is moving vertically there is no acceleration

 

[ideasrule]

Yes, that's right

 

[tiny-tim]

Hi tigerwoods99!

(just got up :zzz: …)

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

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

mu = mg/9.8 * u

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

(oh, if only everybody had a Mac instead of a PC, with a sensible keyboard! )

have a mu … µ

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

but because the object is moving vertically there is no acceleration

(try using the X₂ tag just above the Reply box )

Are you confusing the vertical and horizontal accelerations?

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

Horizontally, F_net = µmg, so ma = µmg.

 

[tigerwoods99]

thanks i got it!