Finding speed of a object on slope

[thewestbrew]

Homework Statement

A 42.5kg person is going down a hill sloped at 42.0 degrees.

The coefficient of kinetic friction between the snow and skis is 0.180.

How fast is the skier going 4.56 seconds after starting from rest?

Homework Equations

ffkmax=coefficient of kinetic friction*Normal Force

fnet=mass*a

mg*cos(42)

mg*sin(42)

The Attempt at a Solution

mg=425

mg*cos(42)=315.8

Fn=mg*cos

ffkmax=(.18)(315.8)=56.8

ffkmax=netforce (no idea from this point on)

fnet=ma 56.8/m=a 56.8N/42.5kg= 1.33ms/^2

V=Vo+a*t

V=(0)+(1.33)(4.56)=6.06m/s

 

[Andrew Mason]

Homework Statement

A 42.5kg person is going down a hill sloped at 42.0 degrees.

The coefficient of kinetic friction between the snow and skis is 0.180.

How fast is the skier going 4.56 seconds after starting from rest?

Homework Equations

ffkmax=coefficient of kinetic friction*Normal Force

fnet=mass*a

mg*cos(42)

mg*sin(42)

The Attempt at a Solution

mg=425

mg*cos(42)=315.8

Fn=mg*cos

ffkmax=(.18)(315.8)=56.8

ffkmax=netforce (no idea from this point on)

fnet=ma 56.8/m=a 56.8N/42.5kg= 1.33ms/^2

V=Vo+a*t

V=(0)+(1.33)(4.56)=6.06m/s

What are the forces acting on the skier? What are the directions of these forces?

You seem to be saying that the friction force is the net force that is accelerating the skier down the hill!

AM