# Newron's 2nd Law on a plane

#### flatline4520

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

A sled moves on the flat surface of a hill. The hill makes an angle θ with the horizontal, and is left orientated.

The hill: θ = 60°; left oriented; μk = 0.1.

The sled: mass = 20 kg; starting location = 20 m from the bottom of the hill;
initial speed = 0 m/s .

http://img26.imageshack.us/img26/1756/blockoninclineleft.jpg [Broken]

Solve:
1. (Fnet)x immediately after motion begins, (N)
2. (Fnet)x, sliding down the hill, (N)
2. Relevant equations

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

Fn = mg cos θ

Fn = 98

3. The attempt at a solution

I draw an FBD and every time I solve for (Fnet)x I get:

(Fnet)x = -mg sin θ - μk*Fn

(Fnet)x = 59.742 - 9.8

(Fnet)x = 49.9 N

Every time I input that as my answer it says it's wrong. Also why is it that the kinetic friction force is a positive 9.8. Doesn't the friction go in the opposite direction of the motion?

Thanks for help

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#### djeitnstine

Gold Member
Welcome to PF.

If a positive friction indicates that to the right is positive and to the left is negative

#### flatline4520

Hi

Thanks that cleared up the friction.. Do you know what I'm doing wrong with the (Fnet)x ?

Thanks for the help

#### slmg_2006

According to my calculations, mg sin(60) (9.8*20*sin(60)) doesn't come out to 59.742...

"Newron's 2nd Law on a plane"

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