How Does Newton's Second Law Apply on an Inclined Plane?

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Homework Help Overview

The discussion revolves around the application of Newton's Second Law to a sled moving down an inclined plane. The problem involves calculating net forces acting on the sled, considering factors such as gravitational force, normal force, and kinetic friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the net force acting on the sled using free body diagrams and equations of motion. They express confusion regarding the sign of the kinetic friction force and its direction relative to the sled's motion.

Discussion Status

Participants are actively engaging with the original poster's calculations and clarifying concepts related to friction. Some guidance has been provided regarding the interpretation of positive and negative directions in the context of forces.

Contextual Notes

There is a noted discrepancy in the calculations of gravitational force components, specifically regarding the value derived from mg sin(60). The discussion also highlights the importance of correctly identifying the direction of forces in the problem setup.

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Homework Statement





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

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

Homework Equations



<br /> \vec{F}_{net} = \Sigma \vec{F} = m \vec{a}<br />Fn = mg cos θ

Fn = 98

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:
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Welcome to PF.

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

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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