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Translational Speed 
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#1
Mar1207, 09:44 PM

P: 5

A bowling ball encounters a 0.760m 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. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.



#2
Mar1307, 07:11 AM

Emeritus
Sci Advisor
PF Gold
P: 4,981

You really need to show some working before you can receive any help. As a hint you can think about the conservation of energy.



#3
May507, 03:12 PM

P: 4




#4
May507, 03:23 PM

P: 4

Translational Speed
1/2(MKG)(7.880)^2+(MKG)(9.8)(.76M)=1/2MKG(V)^2+MKG(9.8)(X) 


#5
May507, 03:25 PM

P: 4




#6
May507, 03:33 PM

HW Helper
P: 1,540

You're on the right track, but you mixed up your gravitational potential energy terms a bit. If you take the bottom of the rise as your reference point, what is the gravitational potential energy of the ball before it goes up the rise?



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