Please check my work--rolling ball?

I don't think there are two forces on the ball at the point of contact as you indicate. I would think the only effective force acting on the ball is kinetic friction.

Although the algebra follows through, I think the error of your reasoning is replacing a/r for α on the third line. I think this because, at that point in time, the ball is not rolling smoothly so a = αr does not apply. This would make it impossible to solve for a supposing there were two forces.

Here is my solution:

(I am assuming v0 is in the positive direction.)

a)

F(net) = ma

- f(kinetic) = ma

- μmg = ma

a = - μg

b)

τ(net) = Iα

Fr = [(2/5)mr^2]α

μmgr = (2/5)mr^2α

α = (5/2)μg/r

I left out the sign b/c it really depends on the given orientation.

c)

v(center of mass) = ωr

v(i) + at = (ω(i) + αt)r

v0 - μgt = (5/2)μgt

t = (2/7)v0/(μg)

d)

d = v(i)t + (1/2)at^2

d = v0[(2/7)v0/(μg)] + (1/2)(- μg)[(2/7)v0/(μg)]^2

d = (12/49)v0^2/(μg)

e)

v(f) = v(i) + at

v = v0 +(- μg)[(2/7)v0/(μg)]

v = (5/7)v0

---------------------

KE(i) = (1/2)mv0^2

KE(f) = (1/2)mv^2 + (1/2)Iω^2

= (1/2)m[(5/7)v0]^2 + (1/2)[(2/5)mr^2][(5/7)v0/r]^2

= (5/14)mv0^2

ΔKE = - (1/7)mv0^2

This loss in KE must be equal to the work done by friction plus the work done by the torque. That is

ΔKE = W(friction) + W(torque)

W(friction) = Fd = (μmg)[(12/49)v0^2/(μg)]

= - (12/49)mv0^2

Since energy was lost to the surroundings the work due to friction is negative work.

W(torque) = τθ

θ = ω(i)t + (1/2)αt^2 = (1/2)[(5/2)μg/r][(2/7)v0/(μg)]^2

= (5/49)v0^2/(rμg)

W(torque) = (μmgr)[(5/49)v0^2/(rμg)] = (5/49)mv0^2

Since rotational kinetic energy was gained by the ball due to the torque, the work done by the torque is positive work.

Check:

ΔKE = W(friction) + W(torque)

- (1/7)mv0^2 = - (12/49)mv0^2 + (5/49)mv0^2

Sorry, kiddo. My eyes are too bad to read such tiny print. How can I enlarge it? I'd be happy to look at it for you.

-Fred