Kinetic Energy of a Particle in a semi-spherical bowl

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The problem involves calculating the kinetic energy of a 410 g particle in a semi-spherical bowl after being released from rest. The particle's speed at point B is given as 2.6 m/s, and the relevant equation for kinetic energy is KE = (1/2)mv^2. The initial calculation of kinetic energy resulted in 1385.8, but there was a reminder to work with units to avoid mistakes, as joules are defined as N⋅m. Correct unit handling is crucial for accurate results in physics problems.
Joshua Lee
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Homework Statement


A 410 g particle in a semi-spherical bowl of radius 0.9 m is released from rest at point A at the level of the center of the bowl, and the surface of the bowl is rough. The speed of the particle at B is 2.6 m/s. The acceleration of gravity is 9.8 m/s 2 . What is its kinetic energy at B? Answer in units of J.

Homework Equations


KE= (1/2)mv^2

The Attempt at a Solution


KE= (1/2)(410)(2.6)^2= 1385.8
 
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You should work with units, that would have helped you to spot your mistake here.
 
Remember that a joule is equal to 1 N⋅m
 

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