Newtons Laws on a sliding box

  • #1
tigerwoods99
99
0

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



//


ANY HELP WOULD BE MUCH APPRECIATED
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,838
256
Hi tigerwoods99! :smile:

Use the work-energy theorem …

work done = change in energy :wink:
 
  • #3
tigerwoods99
99
0
sorry, I am not familiar with this theorem
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,838
256
sorry, I am not familiar with this theorem

oops! :redface:

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. :smile:
 
  • #5
tigerwoods99
99
0
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
ideasrule
Homework Helper
2,284
0
If a=Fnet/m, Fnet=Fnormal*mu, and Fnormal=mg, then a=?
 
  • #7
tigerwoods99
99
0
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
ideasrule
Homework Helper
2,284
0
Just assume the mass is m and try it. You'll find that m cancels out.
 
  • #9
tigerwoods99
99
0
A = Fnet/m
A = (Fnormal *mu)/m

A = (Fnormal *u) ?
So how would I find the FN
 
  • #10
ideasrule
Homework Helper
2,284
0
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?
 
  • #11
tigerwoods99
99
0
mu = mg/9.8 * u
 
  • #12
ideasrule
Homework Helper
2,284
0
"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
tigerwoods99
99
0
thats what i thought, but wasn't sure becuase i am used to seeing it as just u
 
  • #14
tigerwoods99
99
0
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
ideasrule
Homework Helper
2,284
0
Yes, that's right
 
  • #16
tiny-tim
Science Advisor
Homework Helper
25,838
256
Hi tigerwoods99! :smile:

(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 becuase i am used to seeing it as just u

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

have a mu … µ :wink:
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

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

Are you confusing the vertical and horizontal accelerations?

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

Horizontally, Fnet = µmg, so ma = µmg. :smile:
 
  • #17
tigerwoods99
99
0
thanks i got it!
 

Suggested for: Newtons Laws on a sliding box

Replies
20
Views
318
Replies
17
Views
580
Replies
37
Views
985
  • Last Post
Replies
7
Views
349
Replies
5
Views
610
Replies
8
Views
379
Replies
15
Views
577
Replies
7
Views
510
Replies
73
Views
2K
Top