Posted by Syed on Monday, April 25, 2011 at 5:43am.
The weight is
F = M g = 1.96 N
The equilibrium deflection is
Xe = F/k = 3.56*10^-2 m = 3.56 cm.
It will fall to twice that deflection.
Spring and gravitational potential energy, and kinetic energy, are zero when it is initially dropped
(a) The initial total energy, relative to the initial position, is zero.
After falling x = 0.015 m,
GravPE + SpringPE + KE = 0
-M g x + (1/2)kx^2 + (M/2) V^2 = 0
Solve for V.
(b) For max deflection, compute the other location where KE = 0
(1/2)kX^2 = M g X
X = 2 Mg/k, twice the equilibrium deflection
ETi=ETf
Ek+Eg+Ee =Ek+Eg+Ee
Intial/final speed is zero, and having a ref point from the bottom, hf is zero and xi will be zero since in the beging the spring is at equiibrim. Finally ur hi=xf, since the strech will be the distaince traveled.
From this info and the equation we can get:
Eg+Ee= Eg+Ee
mghi+0.5(k)(xi)^2 = mghf+0.5(k)(xf)^2
mghi=0.5(k)(hi)^2
mg=0.5(k)(hi)
mg/0.5(k)=hi
sub in values to get
(0.20)(9.80)/(0.5)(55)=hi
0.071 meters= hi
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