Answer:
[tex]\Delta =2.88\times 10^{-2} J[/tex]
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
[tex]\Delta = \dfrac{1}{2}m(v_f^2-v_i^2)[/tex]
[tex]\Delta = \dfrac{1}{2}\times 0.1 \times (2^2 - 1.85^2)[/tex]
[tex]\Delta = \dfrac{1}{2}\times 0.1 \times (2^2 - 1.85^2)[/tex]
[tex]\Delta = 0.0288 J[/tex]
[tex]\Delta =2.88\times 10^{-2} J[/tex]
