Answer:
The maximum length during the motion is [tex]L_{max} = 1.45m[/tex]
Explanation:
From the question we are told that
      The mass  is  [tex]m =0.5 kg[/tex]
      The vertical spring  length is  [tex]L = 1.10m[/tex]
      The unstretched  length is  [tex]L_{un} = 1.30m[/tex]
     The initial speed is [tex]v_i = 1.3m/s[/tex]
     The new length of the spring [tex]L_{new} = 1.30 m[/tex]
The spring constant k is mathematically represented as
              [tex]k = -\frac{F}{y}[/tex]
Where F is the force applied  [tex]= m * g = 0.5 * 9.8=4.9N[/tex]
      y is the difference in weight which is  [tex]=1.10-0.50=0.6m[/tex]
The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)
  Now  substituting values accordingly
          [tex]k = \frac{4.9}{0.6}[/tex]
            [tex]= 8.17 N/m[/tex]
The  elastic potential energy is given as [tex]E_{PE} = \frac{1}{2} k D^2[/tex]
 where D is this the is the displacement Â
Since Energy is conserved the total elastic potential energy would be
       [tex]E_T = initial \ elastic\ potential \ energy + kinetic \ energy[/tex]
      [tex]E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2[/tex]
Substituting value accordingly
        [tex]\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2[/tex]
        [tex]4.085 * D_{max}^2 = 3.69[/tex]
         [tex]D^2_{max} = 0.9033[/tex]
        [tex]D_{max} = 0.950m[/tex]
So to obtain total length we would add the unstretched length
 So we have
         [tex]L_{max} = 0.950 + 0.5 = 1.45m[/tex]
               Â
       Â
       Â
        Â
          Â
Answer:
Explanation:
The weight of the spring is equal to the restoring force of the spring.
F = mg = k× ΔL
ΔL = change in length of the spring when the mass was first hung on the spring.
Where
Given Lo = 0.5m
L = 1.10m
ΔL = 1.10-0.5 = 0.6m
m = 0.5kg
g = 9.8m/s²
k = mg/ΔL = 0.5×9.8/0.6 = 8.2 N/m
Required to find Lmax
Lmax = Lo + A
A = Amplitude of the oscillation
A = √(ΔLo² + Vo²/ωo²)
ΔLo = 1.30 – 0.5 = 0.8m
Vo = 1.30m/s = initial upward velocity
ωo = initial angular frequency.
ωo = √(k/m) = √(8.2/0.5) = 4.05rad/s
A = √(ΔLo² + Vo²/ωo²)
A = √(0.8² + 1.30²/4.05²) = 0.86m
L = 0.5 + 0.86 = 1.36m
