Question: A mass of 0.3 Kghangs motionless from a vertical spring whose length is 0.8 mand whose unstretched length is 0.65 m. Next the mass is pulled down so the spring has a length of 0.9 mand is given an initial speed upward of 1.2 m/s. What is the maximum length of the spring during the following motion? What approximations or simplifying assumptions did you make?

The term displacement refers to a shift in an object's location. It's a vector quantity with a magnitude and a direction. It's depicted as an arrow pointing from the beginning place to the destination.

The force in a spring is given by:

$F=kx$

Here kis spring constant and xis the distance from the equilibrium point.

Let$l_1$is the unstretched length of the spring and$l_2$is the length of the vertical spring.

Substitute$x=l_2-l_1$into the formula of spring force.

$F=k\left(l_2-l_1\right)$

The downward force will be $F=mg$, where m is the mass, g is the acceleration due to gravity whose value is $9.8\mathrm{m}/{\mathrm{s}}^2$.

Substitute$F=mg$into the obtained formula and solve for k.

$$\begin{gathered} mg=k\left(l_2-l_1\right) \\ k=\frac{mg}{l_2-l_1} \end{gathered}$$

Substitute m = 0.3 kg, $g=9.8\mathrm{m}/{\mathrm{s}}^2$, $l_2=0.8\mathrm{m}$and $l_1=0.65\mathrm{m}$into the obtained result.

$$\begin{gathered} k=\frac{0.3\mathrm{kg}(9.8\mathrm{m}/{\mathrm{s}}^2)}{0.8\mathrm{m}-0.65\mathrm{m}} \\ =19.6\mathrm{N}/\mathrm{m} \end{gathered}$$

Therefore, the value of spring constant is 19.6 N/m.

Let$v_f$ is the final speed,$v_j$is the initial speed,$d_m$is the maximum displacement,$l_s$is the stretched length and$l_1$is the unstretched length.

Apply the conservation of energy to find the maximum length$l_m$.

$$\begin{gathered} E_j=E_f \\ K{E}_f+P{E}_f=K{E}_j+P{E}_j \\ \frac{1}{2}m{v}_f^2+\frac{1}{2}k{d}_m^2=\frac{1}{2}m{v}_j^2+\frac{1}{2}k{\left(l_s=l_1\right)}^2 \end{gathered}$$

Substitute m = 0.3 kg, $v_f=0$, k = 19.6 N/m, $v_i=1.2m/s^2$, $g=9.8m/s^2$, $l_s=0.9m$and $l_1=0.65m$into the obtained result and solve for

$$\begin{gathered} 0+\frac{1}{2}(19.6){\left(d_m\right)}^2=\frac{1}{2}(0.3){(1.2)}^2+\frac{1}{2}(19.6){(0.9-0.65)}^2 \\ \frac{1}{2}(19.6){\left(d_m\right)}^2=0.216+0.6125 \\ {\left(d_m\right)}^2=\frac{0.8285}{9.8} \\ {d}_m=0.29m \end{gathered}$$

Therefore, the maximum displacement is 0.29 m.

Add the value of the maximum displacement,$d_m$and the unstretched length$l_1$to get the maximum length.

$$\begin{gathered} l_{MAX}=d_m+l_1 \\ =0.29\mathrm{m}+0.65\mathrm{m} \\ =0.94\mathrm{m} \end{gathered}$$

Therefore, the maximum length is 0.94 m.