Overcoming Friction

Kurtz · 2007-10-02 12:39:18

I don't know what formula I should be using for this equation. :|

A girl on a sled coasts down a hill. Her speed is 7.0 m/s when she reaches level ground at the bottom. The coefficient of kinetic friction between the sled's runners and the hard, icy snow is 0.050, and the girl and sled together weigh 645 N. How far does the sled travel on the level ground before coming to rest?

JaneFairfax · 2007-10-02 13:27:34

The retarding force is 0.050 × 645 = 32.25 N. The mass of the girl & sled is 645 ∕ 9.8 ≈ 65.82 kg. Hence the deceleration is 32.25 ∕ 65.82 = 0.49 ms[sup]−2[/sup].

Use the equation

to find s.

Kurtz · 2007-10-02 13:31:45

Thank you so much for helping me.

John E. Franklin · 2007-10-02 13:36:42

Forget the 645 Newtons, that cancels out with the mass during deceleration.
(Someone's not gonna like that statement, but you might have to work
that part out the right way, sorry, can't remember)
The speed (V) of the sled with respect to time (t) is probably (if I remember right):
V = 7 - (0.050)(9.8)t, where time (t) = 0 when she reaches level ground.
Inorder to get distance travelled, you can integrate with respect to time (t).
DIST = 7t - (1/2)(0.050)(9.8)t^2
First find when Velocity (V) = 0
0 = V = 7 - (0.050)(9.8)t
t = 14 and 2/7ths seconds
Plug 14.28 number into time (t) in DISTance formula.
DIST = 100 - 50 meters = 50 meters I guess. Hope I did it right.
Now the force in Newtons (N) = F = ma = mg = 9.8(mass)
There's some formulas you will need to do this right.

JaneFairfax · 2007-10-02 22:21:47

Youre right in that the deceleration does not depend on the weight or mass of the girl and sled. I didnt know this before but I can see why now.

But why bring time into the equation? Just use the time-independent equation of motion I gave in my post above.

Last edited by JaneFairfax (2007-10-03 06:43:02)

John E. Franklin · 2007-10-03 04:07:37

Since it takes me ten minutes to do my shabby posts sometimes when I'm interrupted at home, you can conclude from the times on the posts, that I was unaware of your post as I was still posting at the time. But also, I love to do things 50 different ways since my memory is so bad, I'm hoping that one of the ways will actually get stuck in my head, or somehow, some connection will be found that makes it easier for me to do the next time.

mathsyperson · 2007-10-17 05:55:20

This way might involve less work.

Energy = Force x Distance, ∴ Distance = Energy/Force.

E = (1/2)mv²; F = μmg, ∴ D = (1/2)v²/μg = 50 metres.

Math Is Fun Forum

#1 2007-10-02 12:39:18

Overcoming Friction

#2 2007-10-02 13:27:34

Re: Overcoming Friction

#3 2007-10-02 13:31:45

Re: Overcoming Friction

#4 2007-10-02 13:36:42

Re: Overcoming Friction

#5 2007-10-02 22:21:47

Re: Overcoming Friction

#6 2007-10-03 04:07:37

Re: Overcoming Friction

#7 2007-10-17 05:55:20

Re: Overcoming Friction

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