Re: Work-Energy or Work-Kinetic Energy??

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Fri, 7 Nov 1997, Rauber, Joel              Phys wrote:

> My solution to the first part of part (a) is:
>
> x(T) = -(c*m^2/k^2)SIN(SQRT(k/m)*T) + (c*m/k)*T
>
> Where I assumed the initial conditions are that the initial speed is zero 
> and the initial location of the mass is the relaxed position of the spring.
>
> and then proceeded to solve via Newton's 2nd law
> x'' = - (k/m)*x + (c/m)*t
>
> Consequently the PE in the spring is
> 0.5*k*x(T)^2
>
> and is different from what Leigh wrote.
>
> Disclaimer: I worked this out quickly, and make no warrantees regarding its 
> validity.  Any use of this solution in a life threatening situation is 
> strictly unadvised

Joel,

I don't follow the above, (but then my life is not in immediate danger
either.)  Since the spring is massless, the force it applies to the
particle (which is also the net force on the particle) is the same as the
force applied to the spring's free end. It is a simple matter to integrate
the resulting acceleration of the particle twice and find that, at time T,
the particle has moved a distance

  d(T) = (1/6)(C/m)T^3

This is, however, irrelevant, to part a which simply asks for the elastic
potential energy.  Since kx = CT, the potential energy is (CT)^2/2k as
Leigh found.

John
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