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Spring, Mass, and Viscous Friction
Figure 112 shows a system in which a spring is extended by applying a force F to the mass M starting from rest. Let v = the velocity at which the spring is extended, then
where RF = constant of viscous friction, newton seconds per meter
Differentiation of Eq. 157 with respect to time yields
The relationship
is a solution that satisfies Eq. 158 and upon substitution of Eq. 159 in Eq. 158 there results
Equation 160 divided by ε^{mt} yields
or
where j = . Since Eq. 157 is a second order differential equation, two constants of integration are involved in the general solution, and the velocity is expressed by
To evaluate the constants of integration A_{1} and A_{2} let t = 0, Now at t = 0, v = 0 as the mass is started from rest. Furthermore, if
must both be zero at t = 0. Therefore
So
From Eqs. 163 and 165 we have
Substitution of Eq. 166 in Eq. 162 yields
Frictional losses The force required to overcome frictional resistance is
and the power converted into heat, i.e., heat expended in overcoming friction, is
The energy converted into heat through friction is
The same relationship can be derived in a simpler manner on the basis of Eq. 13. The gain in reversible energy is the energy stored in the spring, whereas the irreversible energy is in the form of heat, some of which raises the temperature of part of the system, thus representing a gain in irreversible energy; the remainder of the heat is dissipated to media surrounding the system. When Eq, 13 is applied to this case the following relationship is valid
Accordingly, for this situation Eq. 13 can be reduced to
where X = F/S so that the total energy input
The gain in reversible energy is the energy stored in the spring
Then from Eq. 172 the irreversible energy is
The final energy stored in the spring is also F^{2}/2S. Hence the total energy input to the system is F^{2}/S. Onehalf of this is converted into heat.


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