Chapter 7 - Work and Kinetic Energy - Problems and Conceptual Exercises - Page 213: 64

$21.6mJ$

First of all, we find the energy of a light bulb as: $W_{bulb}=Pt$ We plug in the known values to obtain: $W_{bulb}=(60W)(24h\times 3600s/h)$ $W_{bulb}=5.18\times 10^6J$ Now, we can find the required energy to run an atomic clock for one day: $W_{atoms}=\frac{W_bulb}{240\times 10^6}$ $W_{atoms}=\frac{5.18\times 10^6}{240\times 10^6}$ $W_{atoms}=21.6mJ$