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My question is:

A solid sphere of mass 2.5 kg and radius R =0.20 m is initially rolling along the horizontal without slipping with a center of mass speed V=6.0 m/s

1) if the sphere then rolls without slipping up a fixed ramp(25 deg) and a coefficient of static friction of .35, how far X(in meters) up the ramp does it roll?

Anyway I think I figured this one out correctly, but Im not too certain what to do with the friction. Do I just add it in to the final answer or what?

my work:

Ei=1/2 Mv^2 + 1/2 I(omega)^2

Ef=mgh

Ef=Ei

becomes 1/2 v^2 + 1/5 v^2 = gh (cancel all mass)

becomes h=(v^2/2g)+(v^2/5g)

end up with a height of 2.56 meters up the ramp

2)

Now I am to find the sphere's center of mass acceleration up the ramp

I used a(cm)=-a(t)=-r(alpha)

fs=(I*a(cm))/r^2

a(cm) = gsin(theta)-I*(a(cm)/mr^2)

a(cm) = gsin(theta)/(1+(I/mr^2))

a(cm)=2.961 m/s^2

3)

and finally I am to calculate the static friction acting on the sphere as it rolls up the ramp

I used the same formula as above and plugged in the number a(cm) and of course I got the same answer of 2.961 N

Any help on this problem would be appreciated.

Thanks in advance

M2k