A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.50 m/s. How much heat energy is dissipated to the environment?

The energy lost to the environment is 0.461 J

Explanation:

Given;

mass of the magnet, m = 0.2 kg

height of fall, h = 35 cm = 0.35 m

initial speed of the magnet, u = 0

final speed of the magnet, v = 1.5 m/s

Initial energy of the magnet is given by;

E₁ = P.E₁ + K.E₁

E₁ = mgh₁ + ¹/₂mu²

E₁ = (0.2 x 9.8 x 0.35) + ¹/₂(0.2)(0)²

E₁ = 0.686 J

Final energy of the magnet as it emerges from the tube is given by;

E₂ = mgh₂ +  ¹/₂mv²

E₂ = (0.2 x 9.8 x 0) + ¹/₂(0.2)(1.5)²

E₂ = 0 + 0.225 J

The energy lost to the environment is given by;

E = E₂ - E₁

E = 0.225 J - 0.686 J

E = -0.461 J (negative sign indicates lost energy to the environment)

Therefore, the energy lost to the environment is 0.461 J