Aspool of thin wire (with inner radius r = 0.45 m, outer radius r = 0.60 m, and moment of inertia icm = 0.8208 kg·m2) pivots on a shaft. the wire is pulled down by a mass m = 1.300 kg. after falling a distance d = 0.500 m, starting from rest, the mass has a speed of v = 63.1 cm/s. calculate the energy lost to friction during that time.

Explanation:

The energy lost can be model by the energy equation in potential and kinetic energy in each step of the motion

Given: ,,,,

Kinetic energy:

,

Energy lost :

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j