A 10 cm diameter iron ball having a mass of 41 kg is dropped

A 10 cm diameter iron ball having a mass of 4.1 kg is dropped from a height of 5 km above the Earth\'s surface. When the ball reaches a height of 3 km, the vertical velocity of the ball is a constant V = 150 m/s. The Internal Energy of the ball is U = 540 kJ. What is the Total Energy of the ball relative to an inertial frame of reference having the origin attached to the Earth? What is the Total Energy of the ball if the frame of reference origin is coincident with the ball\'s center and moves with the ball as it falls at constant velocity?

Solution

A)Total energy=internal +potential+kinetic energies

Internal+mgh+mv^2÷2

150+(4.1*9.81*3)+(4.1*(150)^2÷(2*1000))

708kJ.

B) Inertial frame of reference is center of sphere=>radius is =0 & height = 0 ,i.e only internal energy is total energy.

Total energy=54kJ