Question

A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?

Answer 1

h = 2755102 m = 2755.102 km

Step-by-step explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

`mgh = Pt\\\\h = (Pt)/(mg)`

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

`h = ((75\ W)(7200\ s))/((0.02\ kg)(9.8\ m/s^2))`

h = 2755102 m = 2755.102 km