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## Homework Statement

Calculate the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20.

## Homework Equations

Fnet= ma

(V2)^2 = (V1)^2 + 2a(d2-d1)

Ff= uN

W= mg

## The Attempt at a Solution

I'll use @ for theta, and u for mu. Fa= force applied to the sled, Ff=force of friction

With N lying on the positive Y axis and Weight in the x direction lying on the positive x axis, I set Wy=wcos@ and Wx= wsin@, so N=wcos@.

I tried summation of forces:

xFnet = Wx + Fa - Ff = ma

Fa= Ff - Wx + ma

Fa = uN - wsin@ + ma

Fa = u(wcos@) - wsin@ + ma

Fa = umgcos@ - mgsin@ + ma

Fa = m(ugcos@ - gsin@ + a)

I then tried to solve for a using (V2)^2 = (V1)^2 + 2a(d2-d1)

Since the starting velocity was 0 and the starting distance is 0,

A = (V2)^2 / (2d2)

I converted 60 km/hr to m/s, to get 16.66 m/s.

A= (16.66^2)/ 2(75) = 1.85 m/s^2

Plugging that in...

Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))

I got Fa = 20.36 N, though that isn't correct. I'm not sure what I did wrong.