Physics please check
posted by Mary .
A bowling ball encounters a 0.76 m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 3.60 m/s at the bottom of the rise, find the translational speed at the top.
KE = 1/2(2/5)MR^2(V/R)^2= 1/5MV^2
KE = 1/5mv^2  mgh
the mass as stated in the question is uniformly so it can be deleted from the above formula.
KE = 1/5v^2  gh
KE = 1/5(3.60)^2  (9.81 x 0.76)
KE = 2.592  7.4566
KE(final)= 4.8636
Am I right so far? If I am how do I calculate the final speed?
You have only treated the rotational part of the kinetic energy. You also must include the (1/2)MV^2 "translational" part. The total KE is (7/10)M V^2
You also have a problem with signs. The kinetic energy cannot be negative. The sum of potential and kinetic energies is constant. So the decrease in total KE as it goes to the top of the ramp equals the increase in potential energy, MgH.
(7/10)[Vo^2  V1^2) = g H
Vo is the initial velocity. Solve for V1
I would like to say I appreciate your help with my homework!
I did the problem according to the formula you provided and the answer is incorrect. Can you please tell me where I went wrong.
I may have misinterpret the formula. This is what I came up with:
7/10 (V0^2  V1^2)= gh
7/10((3.6)^2  V1^2)= gh
7/10 (12.96  V1^2)= 9.81 X 0.76
9.072  V1^2= 7.4556
V1^2= 9.072  7.4556
V1^2= 1.6164
V1= square root 1.6164
V1= 1.2714
DrWLS is not on right now but one error I see you you didn't multiply V1 by 0.7. I've marked it below.
7/10 (V0^2  V1^2)= gh
7/10((3.6)^2  V1^2)= gh
7/10 (12.96  V1^2)= 9.81 X 0.76
9.072  V1^2= 7.4556 From the previous step, the V1^2 must be 0.7V1^2
V1^2= 9.072  7.4556
V1^2= 1.6164
V1= square root 1.6164
V1= 1.2714
Thank you so much!!!!!
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