Translational Speed of a bowling ball

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SUMMARY

The discussion focuses on calculating the translational speed of a bowling ball at the top of a 0.760-meter vertical rise, given its speed of 7.88 m/s at the bottom. The conservation of energy principle is applied, utilizing the equation kE1 + PE1 = kE2 + PE2. The kinetic energy (kE) and potential energy (PE) are calculated to find the speed at the top, emphasizing the need to correctly set up the gravitational potential energy terms. Participants clarify the equation setup and address common mistakes in energy calculations.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with kinetic and potential energy equations
  • Basic knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to manipulate algebraic equations
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  • Review the conservation of energy in mechanical systems
  • Practice problems involving kinetic and potential energy calculations
  • Learn about the effects of friction on energy conservation
  • Explore real-world applications of energy conservation in sports physics
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Physics students, educators, and anyone interested in understanding the dynamics of motion and energy conservation in sports contexts.

Kelschul
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A bowling ball encounters a 0.760-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. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.
 
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You really need to show some working before you can receive any help. As a hint you can think about the conservation of energy.
 
Kelschul said:
A bowling ball encounters a 0.760-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. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.
HOW IS THE EQUATION SETUP FOR THIS PROBLEM
 
Kelschul said:
A bowling ball encounters a 0.760-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. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.

kE1+PE1=KE2+PE2
1/2(MKG)(7.880)^2+(MKG)(9.8)(.76M)=1/2MKG(V)^2+MKG(9.8)(X)
 
despanie said:
kE1+PE1=KE2+PE2
1/2(MKG)(7.880)^2+(MKG)(9.8)(.76M)=1/2MKG(V)^2+MKG(9.8)(X)

WILL THIS EUATION WORK FOR THE SPEED
 
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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