Question

A frictionless circular ring 1.00 m in diameter is placed flat on the floor, and a 100-g ball is sent around its inside surface with an initial speed of 2.00 m/s. After one full revolution, the speed of the ball measured to be 1.85 m/s. (a) How much energy is converted to internal energy, heating up the ball and the floor, per revolution

Answer 1

Answer:

 $\Delta =2.88\times 10^{-2} J$

Explanation:

given,

diameter of the ring = 1 m

radius = 0.5 m

mass of the ball = 100 g = 0.1 Kg

initial speed of the ball = 1.85 m/s

converted energy = change in kinetic energy

 $\Delta = \dfrac{1}{2}m(v_f^2-v_i^2)$

 $\Delta = \dfrac{1}{2}\times 0.1 \times (2^2 - 1.85^2)$

 $\Delta = \dfrac{1}{2}\times 0.1 \times (2^2 - 1.85^2)$

 $\Delta = 0.0288 J$

 $\Delta =2.88\times 10^{-2} J$