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Re: test problem



1) Define x=0 as the mass location when the spring is relaxed (the mass is removed or is supported "by hand"), and take x as positive upwards.  Note this makes the elastic PE easy to evaluate as simply .5*k*x^2 - the presence of other forces acting on the mass does not change this.  For convenience, also use the same x=0 point as the zero of the gravitational PE.
 
2) The rest (equilibrium) position with the mass released (to the earth's attraction) is x = a.
 
3) Energy balance at two arbitrary positions, x1 and x2 , in the motion yields:
 
.5*k*x1^2 + m*g*x1 + .5*m*v1^2 = .5*k*x2^2 + m*g*x2 + .5*m*v2^2
 
Given:
G1)  With m=1 kg,   a = - .05m    =>  k = -mg/a =9.8/.05 = 196N/m
G2)  The suspended mass is released from rest at x = -.25m
 
Apply Eq 3), letting x1 be the coordinate of the midpoint of the oscillation, and x2 be the position of release from rest:
Numerically:  x1 = a = -.05m;   x2 = -.25m ; v2 = 0 (Also remember k = -mg/a = 196N/m)
 
Eq 3) and algebra then yield v1 = 2.8m/sec
 
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
www.velocity.net/~trebor