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How much work is required to stop an electron (m=9.11×10−31kg) which is moving with a speed of 2.10×106 m/s ?

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Alize Kuhn

Feb 20, 2025

To decrease the electron’s velocity from 2.10 * 10^6 m/s to 0 m/s, the work must be equal to the electron’s initial kinetic energy.KE = ½ * 9.11 * 10^-31 * (2.10 * 10^6)^2 = 2.008755 * 10^-18 JThis is the work.... Show More
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A
Alize Kuhn

Feb 20, 2025

Enough work QE to reduce the KE to zero. So we have KE – QE = 0 and 1/2 mU^2 = QE = .5*9.11E-31*(2.1E6)^2 = 2.01E-18 Joules. ANS.... Show More
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