<p>"A child drags his 60N sled at constant speed up a 15 degree hill. He does so by pulling with a 25N force on a rope attached to the sled. If the rope is inclined 35 degrees to the horizontal, (a) what is the coefficient of kinetic friction between sled and snow? (b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?"</p>

<p>Correct answers: (a) 0.161 (b) 1.01 m/s^2. I would appreciate help in how to get to these answers.</p>

<p>My current work:</p>

<p>a)

Ffric = uN

since Fnety=0, N=W

thus u=Ffric/W<em>sin(15degrees) [sin15 to get component]

next, since speed is constant there must nto be any a in the x

therefore Ffric=Fpull

rewriting... u = Fpull</em>??? / W*sin(15degrees) = 0.161

My big problem is what is Fpull adjusted by to get this to work out? cos(20) seems logic, but it (and all other cos/sin values of 15 and 35 combined/subtracted/etc don't seem to work).</p>

<p>b) Fnet=ma, thus a=(Fweight-Ffriction)/m=(W * ( sin15 - .161*sin(20)) ) / (W/g)

But that produces 1.99 (not 1.01).</p>

<p>Any help would be greatly appreciated.</p>