Question

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?

Short answer

The maximum length obtained for the spring during the motion is $l_{MAX}=0.94\mathrm{m}$.

Step-by-step solution

Definition of the displacement

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.

Finding the value of spring constant

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.

Finding maximum displacement

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.

Finding maximum length

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.