The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s
The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²
The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j
The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J
The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J
Answer: 0.0226 J
