A 20-cm-diameter, 6.0kg bowling ball rolls along a horizontal floor at 4.0 m/s. The ball then rolls up a 2.0-m-long ramp sloped at 30 degrees up from horizontal. Rolling friction is negligible. What is the ball's speed at the top of the ramp?
K(final) + U(final) = K(initial) + U(initial)
I = 2/5 * M * R^2
omega = V/R
The Attempt at a Solution
The ball starts out with all kinetic energy so the above equation becomes k(final) + U(final) = k(initial). I set up the following equation:
1/2*I*omega(final)^2 + 1/2*M*v(final)^2 + M * g * h = 1/2*I*omega(initial)^2 + 1/2*M*v(initial)^2
After substituting in the values for I and omega, the radius and mass cancel out and I am left with:
7/10*v(final)^2 = 7/10*v(initial)^2 - g * h
After doing some simple trig on the slope I found that h = 1m, and solving for v(final) I end up with:
v(final) = 1.41 m/s
Since I don't know the answer to the problem and I get credit on an all or nothing basis, I wanted to come here and have someone verify the validity of my work before I hand it in.