Newron's 2nd Law on a plane

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 . [Broken]

  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
Last edited by a moderator:


Gold Member
Welcome to PF.

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

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

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

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