Question

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the mass is pulled down to where the spring has a length of 1.30 m and given an initial speed upwards of 1.3 m/s. What is the maximum length of the spring during the motion that follows?

Answer 1

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

           The mass  is  $m =0.5 kg$

            The vertical spring  length is  $L = 1.10m$

            The unstretched  length is  $L_{un} = 1.30m$

          The initial speed is $v_i = 1.3m/s$

          The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

                           $k = -\frac{F}{y}$

Where F is the force applied  $= m * g = 0.5 * 9.8=4.9N$

           y is the difference in weight which is   $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    $k = \frac{4.9}{0.6}$

                       $= 8.17 N/m$

The  elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             $E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

            $E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

                $\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

                $4.085 * D_{max}^2 = 3.69$

                 $D^2_{max} = 0.9033$

                $D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

 So we have

                  $L_{max} = 0.950 + 0.5 = 1.45m$

                               

               

               

                 

                     

Answer 2

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