A 1.60 Times 102-g particle on the inside of a smooth hemispherical bowl of radi

Question

A 1.60 Times  102-g particle on the inside of a smooth hemispherical bowl of radius R = 26.0 cm (see the figure below) is released from point A at rest. Its speed at B is 1.60 m/s. What is its kinetic energy at B? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J How much mechanical energy is lost as a result of friction as the particle goes from A to B? Is it possible to determine ? from these results in a simple manner?

Explanation / Answer

a) KEb = 0.5*m*V^2 = 0.5*1.6*10^2*1.6^2 = 204.8 J

b) KEa = m g R = 1.6*10^2*9.8*0.26 = 407.68 J

loss = KEa - KEb = 202.88 J

c) no

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