Conservation of energy of an ice flake

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SUMMARY

The discussion centers on calculating the speed of a 1.90 g ice flake released from the edge of a hemispherical bowl with a radius of 28.0 cm. The participant correctly applies the principle of conservation of energy, equating kinetic energy and gravitational potential energy. The formula derived is V = √(2gR), where g is the acceleration due to gravity. After converting the radius from centimeters to meters, the final speed of the ice flake is confirmed to be 2.344 m/s.

PREREQUISITES
  • Understanding of kinetic energy and gravitational potential energy concepts
  • Familiarity with the formula V = √(2gR)
  • Ability to perform unit conversions, specifically from centimeters to meters
  • Basic knowledge of physics principles related to motion and energy conservation
NEXT STEPS
  • Study the principles of conservation of energy in physics
  • Learn about unit conversions in scientific calculations
  • Explore examples of kinetic and potential energy problems
  • Investigate the effects of friction on energy conservation in physical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation principles, as well as educators looking for practical examples of energy calculations.

lauriecherie
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Homework Statement



A 1.90 g ice flake is released from the ege of a hemispherical bwon whose radius r is 28.0 cm. The flake bowl contact is frictionless. What is the speed of the flake when it reaches the bottom of the bowl.

Homework Equations


I took KINETIC ENERGY = GRAVITATIONAL POTENTION ENERGY. since KINETIC ENERGY = .5*M*V^2, and GRAVITATIONAL POTENTIAL ENERGY = MGH, i canclled the masses and solved for V. That gave me VELOCITY = SQUARE ROOT OF (2*G*R).


The Attempt at a Solution


I got 23.44 for my answer but after looking over my calculations I noticed that the radius is given in cm. Need I convert it to meters? If I do my answer comes out as 2.344 m/s. Is this correct now?
 
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Nevermind. It is correct.
 

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