Chapter 8, Problem 15P

(a)

To determine

The speed of block if the horizontal surface is frictionless.

Answer to Problem 15P

The speed of block if the horizontal surface is frictionless is $0.791\text{m}/\text{s}$ .

Explanation of Solution

Draw the figure fir the system as shown below.

In this system block is pulled to some distance and then released from rest, so string comes back to its normal position.

Write the expression for elastic energy stored in the spring as.

  $W_s=\frac{1}{2}k{x}^2$                                                                                                      (I)

Here, $W_s$ is elastic energy, $k$ is force constant and $x$ is compression in the spring.

Write the expression for work done by the spring on the block.

  $W_s=\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2$                                                                                        (II)

Here, $x_i$ is initial stretch of the spring and $x_f$ is final stretch of the spring whole value is zero as string will come to its normal state.

Write the expression for kinetic energy of the system.

    $K=\frac{1}{2}m{v}^2$                                                                                                 (III)

Here, $K$ is kinetic energy, $v$ is velocity and $m$ is mass.

Write the expression for change in kinetic energy of the system.   

  $\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$                                                                                   (IV)

Here, $\Delta K$ is change in kinetic energy, $v_i$ is initial velocity and $v_f$ is final velocity.

As energy of the system is conserved, therefore work done by the string is equal to change in kinetic energy.

Write the expression for change conservation of energy for the system.

    $\Delta K=W_s$

Substitute  $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ for $\Delta K$ and $\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2$ for $W_s$ in above expression.

  $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2$                                                                    (V)

Value of stretch at final position is zero as string comes to normal state and initial velocity is also zero.

Substitute $0$ for $\frac{1}{2}m{v}_i^2$ and $0$ for $\frac{1}{2}k{x}_f^2$ in equation (V).   

  $\frac{1}{2}m{v}_f^2=\frac{1}{2}k{x}_i^2$

Rearrange the above equation for $v_f$

  $v_f=\sqrt{\frac{k{x}_i^2}{m}}$                                                                                                 (VI)

Conclusion:

Substitute  $\left(500\text{N}/\text{m}\right)$ for $k$, $\left(5.00\times {10}^{-2}\text{m}\right)$ for $x_i$ and $\left(2.00\text{kg}\right)$ for $m$ in equation (VI).

  $\begin{array}{c}{v}_f=\sqrt{\frac{\left(500\text{N}/\text{m}\right){\left(5.00\times {10}^{-2}\text{m}\right)}^2}{\left(2.00\text{kg}\right)}}\\ =\sqrt{\frac{\left(1.25\text{N}\cdot \text{m}\right)}{\left(2.00\text{kg}\right)}}\\ =0.7905\text{m}/\text{s}\\ \simeq 0.791\text{m}/\text{s}\end{array}$

Thus, the speed of block if the horizontal surface is frictionless is $0.791\text{m}/\text{s}$ .

(b)

To determine

The speed of block if there is some friction between block and surface.

Answer to Problem 15P

The speed of block if there is some friction between block and surface is $0.531\text{m}/\text{s}$.

Explanation of Solution

Draw the figure for the system considering some friction as shown below.

Friction force results in an increase in internal energy of the block-surface system.

Write the expression for the normal reaction force experience by the block.

    $n=mg$                                                                                                     (VII)

Here, $n$ is normal reaction and $g$ is acceleration due to gravity.

Write the expression for friction force.

    $f_k={\mu }_{s}n$                                                                                                   (VIII)

Here, $f_k$ is frictional force, ${\mu }_s$ is coefficient of static friction.

Substitute $mg$ for $n$ in equation (VIII)

  $f_k={\mu }_{s}mg$                                                                                                  (IX)

Here, $d$ is the distance by which the box moves when string is stretched which is equal to $x_i$ .

Write the expression for total internal energy.

    $\Delta E_{\mathrm{int}}=f_{k}d$                                                                                                   (X)

Here, $\Delta E_{\mathrm{int}}$ total internal energy stored in the system.

Substitute ${\mu }_{s}mg$ for $f_k$ in equation (X)

  $\Delta E_{\mathrm{int}}={\mu }_{s}mgd$                                                                                             (XI)

Work done by string is total of change in kinetic energy and internal energy.

Write the expression for change conservation of energy for the system.

    $W_s=\Delta K+\Delta E_{\mathrm{int}}$

Rearrange the above equation.

    $\Delta K=W_s-\Delta E_{\mathrm{int}}$                                                                                          (XII)

Substitute $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ for $\Delta K$, $\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2$ for $W_s$ and ${\mu }_{s}mgd$ for $\Delta E_{\mathrm{int}}$ in equation (XII).   

  $\left(\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2\right)=\left(\frac{1}{2}k{x}_i^2-\frac{1}{2}k{x}_f^2\right)-{\mu }_{s}mgd$                                           (XIII)

Value of stretch at final position is zero as string comes to normal state and initial velocity is also zero.

Substitute  $0$ for $\frac{1}{2}m{v}_i^2$, $0$ for  $\frac{1}{2}k{x}_f^2$ and $x_i$ for $d$  in equation (V).      

  $\frac{1}{2}m{v}_f^2=\frac{1}{2}k{x}_i^2-{\mu }_{s}mg{x}_i$

Rearrange above equation for $v_f$.

    $v_f=\sqrt{\frac{2}{m}\left[\left(\frac{1}{2}k{x}_i^2\right)-{\mu }_{s}mg{x}_i\right]}$                                                                  (XIV)

Conclusion:

Substitute  $\left(500\text{N}/\text{m}\right)$ for $k$ , $\left(5.00\times {10}^{-2}\text{m}\right)$ for $x_i$ and $\left(2.00\text{kg}\right)$ for $m$, $\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)$ for $g$ and $0.350$ for ${\mu }_s$  in equation in equation (XIV).

    $\begin{array}{l}{v}_f=\sqrt{\frac{2}{\left(2.00\text{kg}\right)}\left[\begin{array}{l}\left(\frac{\left(500\text{N}/\text{m}\right){\left(5.00\times {10}^{-2}\text{m}\right)}^2}{2}\right)-\\ \left(\left(0.350\right)\left(2.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(5.00\times {10}^{-2}\text{m}\right)\right)\end{array}\right]}\\ =\sqrt{\frac{2}{\left(2.00\text{kg}\right)}\left[\left(0.625\text{N}\cdot \text{m}\right)-\left(0.343\text{N}\cdot \text{m}\right)\right]}\\ =\sqrt{\frac{2\left(0.282\text{N}\cdot \text{m}\right)}{\left(2.00\text{kg}\right)}}\\ =0.531\text{m}/\text{s}\end{array}$

Thus, the speed of block if there is some friction between block and surface is $0.531\text{m}/\text{s}$.