How much work is required to stop an electron $(m = 9.11 \times 10^{-31} \textrm{ kg})$ which is moving with a speed of $1.10 \times 10^6 \textrm{ m/s}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The net work done on this electron is its change in kinetic energy, which is, the final kinetic energy minus the initial. But it comes to a stop so the final kinetic energy is 0. And that means we just have negative one-half m v initial squared for the net work. So that's equal to negative one-half times the mass of an electron—9.11 times 10 to the negative 31 kilograms— times its initial speed of 1.10 times 10 to the 6 meters per second. And we square that and we get negative 5.51 times 10 to the minus 19 joules. And it's a negative work done, partly because that's what the algebra works out to, because we have this minus initial kinetic energy and you can also say that the net work is negative when the speed is being reduced.