How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

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Answer 1

Answer #1

yeah... the question is a little ambiguous. But if the electron is slowed down, you have to do negative work on it.

initial KE = (1/2) m v^2 = (1/2) * 9.11 x 10^-31 * (2100000)^2 = 2.009 x 10^-18 Joules

final KE = 0

work done on electron = final KE - initial KE =

= 0 - 2.009 x 10^-18 J

= - 2.009 x 10^-18 J

Answer 2

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