Question

A 0.20-kg mass is hung from a vertical spring of force constant 55 N/m. When the
spring is released from its unstretched equilibrium position, the mass is allowed to
fall. Use the law of conservation of energy to determine
(a) the speed of the mass after it falls 1.5 cm
(b) the distance the mass will fall before reversing direction

Answer 1

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

Answer 2

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