aldentmainfr.uni ent FULL SCREEN PRINTER VERSION Chapter 18, Problem 24 GO ·BACK

Question

aldentmainfr.uni ent FULL SCREEN PRINTER VERSION Chapter 18, Problem 24 GO ·BACK NEN Your answer is partially correct. Try again. An unstrained horizontal spring has a length of 0.32 m and a spring constant of 250 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.036 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges (a)T Either both charges are positive or both charges are negative. Number 1.13842E-6 the tolerance is +/-2% SHoW MENT G0 TUTORIAL LINK TO TEXT Question Attempts: Unlimited SAVE FOR LATER SUBHIT ANSWER powered by MapleNet olor ! 2000-2018 ohr-Wiley asonalnc. All Rights Reserved. A Division of John wlnA Version 4.24.4.5

Explanation / Answer

here,

untreched length , l = 0.32 m

spring constant , K = 250 N/m

x = 0.036 m

let the magnitude of charges be q

using conservation of energy

the change in potential energy of charges = potential energy stored in the spring

k * q * q * ( 1/x - 1/(l + x) ) = 0.5 * k * x^2

9 * 10^9 * q^2 /( 1/0.32 - 1/(0.32 + 0.036)) = 0.5 * 250 * 0.036^2

solving for q

q = 7.5 * 10^-6 C = 7.5 uC

the magnitude of charges is 7.5 uC

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