Question

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.0400 m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm (Fig. P8.78). Find the maximum distance the frame moves downward from its initial equilibrium position.

Short answer

The maximum distance moved by the frame from its equilibrium position is $\mathbf{0}\mathbf{.}\mathbf{048}\mathbf{\hspace{0.33em}}\mathbf{m}$.

Step-by-step solution

Determination of speed of frame and force constant of spring.Given Data:

The height of putty is$\mathrm{h}=30\hspace{0.33em}\mathrm{cm}=0.3\hspace{0.33em}\mathrm{m}$

The initial stretch of spring is $\mathrm{x}=0.0400\hspace{0.33em}\mathrm{m}$

The mass of frame is: $\mathrm{m}=0.150\mathrm{kg}$

The mass of putty is: $\mathrm{M}=0.200\hspace{0.33em}\mathrm{kg}$

The speed of frame is found by momentum conservation of frame and putty. The maximum distance travelled by frame from its initial equilibrium position is found by energy conservation.

The speed of putty before strike on frame is calculated as:

$v_p=\sqrt{2gh}$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{p}}=\sqrt{\left(2\right(9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2\left)\right(0.3\hspace{0.33em}\mathrm{m})} \\ {\mathrm{v}}_{\mathrm{p}}=2.42\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Apply the momentum conservation for frame and putty.

$\left(\mathrm{m}+\mathrm{M}\right){\mathrm{v}}_{\mathrm{f}}={\mathrm{mv}}_{\mathrm{p}}$

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.150\mathrm{kg}+0.200\mathrm{kg}\right){\mathrm{v}}_{\mathrm{f}}=\left(0.200\mathrm{kg}\right)\left(2.42\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{f}}=1.38\mathrm{m}/\mathrm{s} \end{gathered}$$

The force constant of the spring is calculated as:

$kx=\left(m+M\right)g$

Here, k is the force constant and g is the gravitational acceleration.

$$\begin{gathered} \mathrm{k}\left(0.0400\mathrm{m}\right)=\left(0.150\mathrm{kg}+0.200\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{k}=85.75\mathrm{N}/\mathrm{m} \end{gathered}$$

Determination of maximum distance moved by frame from equilibrium position

The maximum distance moved by the frame from its equilibrium position is given as:

$$\begin{gathered} \frac{1}{2}k{\left(x+I\right)}^2=\frac{1}{2}\left(m+M\right){v_f}^2 \\ k{\left(x+I\right)}^2=\left(m+M\right){v_f}^2 \end{gathered}$$

Here, I is the maximum distance moved by frame.

Substitute all the values in the above equation.

$$\begin{gathered} \left(85.75N/m\right){\left(0.040m+I\right)}^2=\left(0.150kg+0.200kg\right){\left(1.38m/s\right)}^2 \\ 0.040m+I=0.088m \\ I=0.088m \end{gathered}$$

Therefore, the maximum distance moved by the frame from its equilibrium position is 0.048 m.