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



John,
I misinterpreted the set-up of the problem, I interpreted pulling on the
spring to mean pulling on the mass with other end of spring attached to a
rigid support. Thus two forces on the mass. Leigh, has since informed me
that he means pulling on the spring's opposite end, opposite to the side the
mass is on.

I think I'm missing something simple though, as I don't understand what you
wrote below, and I erased the original problem so I can't look at it.

Am I to envision a massless spring of length l_o, with a mass m attached at
one end, the whole system floating in outer space, far from any planets,
stars, etc and other outside influences; then I come along and pull on the
other end with a force proportional to time, i.e. c*t ?
If so then I have some further questions.

Joel
----------
From: John Mallinckrodt
To: QuistO; RAUBERJ; phys-l
Cc: AJMALLINCKRO
Subject: Re: Work-Energy or Work-Kinetic Energy??
Date: Friday, November 07, 1997 11:47AM

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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