A 150 g particle in a semi-spherical bowl of radius 0.8 m is released from rest

Question

A 150 g particle in a semi-spherical bowl of radius 0.8 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 2.9 m/s. The acceleration of gravity is 9.8 m/s^2

What is its kinetic energy at B? Answer in units of J

What is the magnitude of the energy lost due to friction as the particle moves from A to B?

If you could give an in depth answer. I need to be able to understand and reporduce the work.

Answer in units of J

Explanation / Answer

K.E at B=1/2 mv^2=0.63075 J

Energy lost =initial energy-final enegy

initial energy=mgR

final energy=1/2 mv^2

==> Energy lost =mgR-1/2 mv^2=0.15*9.8*0.8-0.63075=0.54525 J

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