Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems: 28a

Answer

$v=x\sqrt {\dfrac {k}{m}}=5.5\times 10^{-2}m\times \sqrt {\dfrac {10\dfrac {N}{m}}{3.8\times 10^{-3}kg}}\approx 2.82\dfrac {m}{s}$

Work Step by Step

Firstly let's calculate spring constant from the graph: $k=\dfrac {\Delta F}{\Delta x}=\dfrac {0,4N}{4\times 10^{-2}m}=10\dfrac {N}{m}$ Lets think cork and spring as a system So initialy this system has only potential energy stored in spring so $E_{tot}=\dfrac {kx^{2}}{2}\left( 1\right) $ When the cork leaves spring when spring is in relaxed position the potential energy of system will be zero so all the potential energy stored in spring will convert to kinetic energy of cork Lets write totat energy equation again (total energy is conserved) $E_{tot}=\dfrac {mv^{2}}{2}\left( 2\right) $ So from (1) and (2) we get $v=x\sqrt {\dfrac {k}{m}}=5.5\times 10^{-2}m\times \sqrt {\dfrac {10\dfrac {N}{m}}{3.8\times 10^{-3}kg}}\approx 2.82\dfrac {m}{s}$
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