Question

A block of mass $\mathit{m}\mathbf{=}\mathbf{2}\mathbf{\text{kg}}$ is attached to a spring of force constant$\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{500}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$ as shown in Figure P8.15. The block is pulled to a position ${\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathbf{5}\mathbf{\text{cm}}$to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if

(a) the horizontal surface is frictionless and

(b) the coefficient of friction between block and surface is ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}$.

Short answer

(a) The speed the block as it passes through equilibrium if the horizontal surface is frictionless is $0.79m/s$.

(b) The speed the block has as it passes through equilibrium if the coefficient of the friction between the block and the surface is ${\mu }_k=0.35$ is $0.53m/s$.

Step-by-step solution

Given data

The mass of the block is$m=2\text{kg}$.

The force constant of the spring is $k=500N/m$.

Extension of the spring is $x_i=5\text{cm}=0.05\text{m.}$

In the second case, the coefficient of the friction between the block and the surface is ${\mu }_k=0.35$.

Spring energy, kinetic energy, and energy to overcome friction

The energy stored in a spring of spring constant $\mathit{k}$ elongated or compressed by a distance $\mathit{x}$ is:

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$ ..... (I)

Here, $\mathit{g}$ is the acceleration due to gravity having a value of $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$.

The kinetic energy of a mass $\mathit{m}$ moving with a velocity $\mathit{v}$ is:

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ ..... (II)

The work done by the friction having the coefficient $\mathit{\mu }$ on a mass $\mathit{m}$ through a distance $\mathit{x}$ is:

${\mathit{W}}_{\mathbf{k}}\mathbf{=}\mathit{\mu }\mathit{m}\mathit{g}\mathit{x}$ ..... (III)

(a) Determining the speed of the block in the absence of friction

From the conservation of energy, the initial spring energy is equal to the final kinetic energy. Thus, from equations (I) and (II), we have:

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}k{x}_i^2 \\ v^2=\frac{k{x}_i^2}{m} \\ v=x_i\sqrt{\frac{k}{m}} \end{gathered}$$

Replace the known values in the above equation, and e have:

$$\begin{gathered} v=0.05m\times \sqrt{\frac{500N/m}{2kg}} \\ =0.05m\times \sqrt{250\times \left(1N\times \frac{1kg\times m/s^2}{1N}\right)\times \left(\frac{1}{1kg\times m}\right)} \\ =0.05m\times 15.8s^{-1} \\ =0.79m/s \end{gathered}$$

Thus, the speed the block as it passes through equilibrium if the horizontal surface is frictionless is $0.79m/s$.

(b) Determining the speed of the block in the presence of friction

In this case, a part of the energy is consumed to overcome friction. Thus, from equations (I), (II), and (III) we have:

$$\begin{gathered} \frac{1}{2}m{v}^2+{\mu }_{k}mg{x}_i=\frac{1}{2}k{x}_i^2 \\ v^2=\frac{k{x}_i^2}{m}-2{\mu }_{k}g{x}_i \\ v=\sqrt{\frac{k{x}_i^2}{m}-2{\mu }_{k}g{x}_i} \end{gathered}$$

Replace the values to get the following.

$$\begin{gathered} v=\sqrt{\frac{500N/m\times {(0.05)}^2}{2kg}-2\times 0.35\times 9.8m/s^2\times 0.05m} \\ =\sqrt{0.625·\left(1N\times \frac{1kgm/s^2}{1N}\right)·\left(1m\right)·\left(\frac{1}{1kg}\right)-0.343m^2/s^2} \\ =\sqrt{0.282m^2/s^2} \\ =0.53m/s \end{gathered}$$

Hence, the speed the block as it passes through equilibrium if the coefficient of the friction between the block and the surface is ${\mu }_k=0.35$ is $0.53m/s$.