A spring with spring constant k = 620 N/mis placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed 25 cm, and a block with a weight of 50 Nis placed (unattached) on the depressed spring. The system is then released from rest. Assume that the gravitational potential energy ${\mathit{U}}_{\mathbf{g}}$of the block is zero at the release point role="math" localid="1661235142508" $\left(y=0\right)$and calculate the kinetic energyof the block forequal to (a) 0, (b) 0.050 m, (c) 0.10 m, (d) 0.15 m, and (e) 0.20 m. Also, (f) how far above its point of release does the block rise?

The spring stiffness is$k=620\mathrm{N}/\mathrm{m}$

The weight of the block is$W=50\mathrm{N}$

The depression of the upper end,$x=25\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.25\mathrm{m}$

We can use conservation of energy in which the summation of kinetic energy, change in potential energy, and change in elastic energy is zero.

Formulae:

The kinetic energy of a spring, $K=\frac{1}{2}{\mathrm{mv}}^2$ (1)

The potential energy of a body, ${\mathrm{U}}_{\mathrm{g}}=\mathrm{mgy}$ (2)

The elastic potential energy of a spring, ${\mathrm{U}}_{\mathrm{e}}=\frac{1}{2}{\mathrm{ky}}^2$ (3)

Using the conservation of mechanical energy, we can write

$$\begin{gathered} {\mathrm{K}}_{\mathrm{i}}+\mathrm{U}={\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{f}} \\ 0+\frac{1}{2}{\mathrm{ky}}_{\mathrm{i}}^2={\mathrm{mv}}_{\mathrm{f}}^2+\frac{1}{2}{\left({\mathrm{y}}_{\mathrm{f}}-{\mathrm{y}}_{\mathrm{i}}\right)}^2+{\mathrm{mgy}}_{\mathrm{f}} \end{gathered}$$

Initial depression of the spring isandis the displacement of the spring from its equilibrium position when the block is at the final position. Thus, the kinetic energy of the block using equations (1) and (3) is written as

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\frac{1}{2}\mathrm{k}\left[{\mathrm{y}}_{\mathrm{i}}^2{\left({\mathrm{y}}_{\mathrm{f}}-{\mathrm{y}}_{\mathrm{i}}\right)}^2\right]-\mathrm{mgy}...............\left(4\right)$

For $y=0$, kinetic energy is zero. Therefore, the kinetic energy is zero.

For$y=0.05\mathrm{m}$ , the kinetic energy of the body using equation (4) is given as:

role="math" localid="1661237580491" $$\begin{gathered} K_f=\frac{1}{2}\left(620\mathrm{N}/\mathrm{m}\right)\left[{\left(0.25\mathrm{m}\right)}^2-{\left(0.050\mathrm{m}-0.25\mathrm{m}\right)}^2\right]-\left(50\mathrm{N}\right)\left(0.050\mathrm{m}\right) \\ =4.48\mathrm{N}·\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =4.48\mathrm{J} \end{gathered}$$

Hence, the value of energy is $4.48\mathrm{J}$.

For$y_f=0.10\mathrm{m}$ , we have the kinetic energy using equation (4) as:

role="math" localid="1661237654360" $$\begin{gathered} K_f=\frac{1}{2}\left(620\mathrm{N}/\mathrm{m}\right)\left[{\left(0.25\mathrm{m}\right)}^2-{\left(0.10\mathrm{m}-0.25\mathrm{m}\right)}^2\right]-\left(50\mathrm{N}\right)\left(0.10\mathrm{m}\right) \\ =7.40\mathrm{N}·\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =7.40\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $7.40\mathrm{J}$.

At, the kinetic energy is given using equation (4) as:

$$\begin{gathered} K_f=\frac{1}{2}\left(620\mathrm{N}/\mathrm{m}\right)\left[{\left(0.25\mathrm{m}\right)}^2-{\left(0.15\mathrm{m}-0.25\mathrm{m}\right)}^2\right]-\left(50\mathrm{N}\right)\left(0.15\mathrm{m}\right) \\ =8.78\mathrm{N}·\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =8.78\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $8.78\mathrm{J}$.

At$y_f=0.20\mathrm{m}$ , the kinetic energy is given using equation (4) as:

$$\begin{gathered} K_f=\frac{1}{2}\left(620\mathrm{N}/\mathrm{m}\right)\left[{\left(0.25\mathrm{m}\right)}^2-{\left(0.15\mathrm{m}-0.25\mathrm{m}\right)}^2\right]-\left(50\mathrm{N}\right)\left(0.15\mathrm{m}\right) \\ =8.60\mathrm{N}·\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =8.60\mathrm{J} \end{gathered}$$

Hence, the value of the energy is $8.60\mathrm{J}$.

The spring returns to its original state once ${\mathrm{y}}_{\mathrm{f}}>{\mathrm{y}}_{\mathrm{i}}$. At this position, the block is detached from the spring. At its maximum height, $\mathrm{K}=0$. Therefore,

$$\begin{gathered} \frac{1}{2}{\mathrm{ky}}_{\mathrm{i}}^2={\mathrm{mgy}}_{\max } \\ {\mathrm{y}}_{\max }=\frac{{\mathrm{ky}}_{\mathrm{i}}^2}{2\mathrm{m}\mathrm{g}} \\ {\mathrm{y}}_{\max }=\frac{\left(620\mathrm{N}/\mathrm{m}\right){\left(0.25\mathrm{m}\right)}^2}{2\left(50\mathrm{N}\right)10\mathrm{m}/{\mathrm{s}}^2} \\ {\mathrm{y}}_{\max }=0.038\mathrm{m} \end{gathered}$$

Hence, the block rise is $0.038\mathrm{m}$.