Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 27 - Early Quantum Theory and Models of the Atom - General Problems - Page 801: 72

Answer

$8.3\times10^{-9}N$.

Work Step by Step

Momentum is conserved. The light leaving the flashlight gains momentum, given by equation 22–9a. Dividing the change in momentum by the elapsed time tells us the force that the flashlight applies to the emitted light. $$\frac{\Delta p}{\Delta t}=\frac{\Delta U}{c\Delta t}=\frac{P}{c}$$ $$=\frac{2.5W}{3.00\times10^8 m/s}=8.3\times10^{-9}N $$ By Newton’s third law, this equals the reaction force exerted upon the flashlight.
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