Chapter 9, Problem 9.87AP

Review. A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0.500-kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is (a) 0, (b) 0.100, and (c) 0.462. Assume the coefficient of static friction is greater than the coefficient of kinetic friction in every case.

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is 0.

Answer to Problem 9.87AP

The maximum velocity of each block is $-0.256\widehat{\text{i}}\text{m}/\text{s}$ and $0.128\widehat{\text{i}}\text{m}/\text{s}$ respectively.

Explanation of Solution

The force constant is $3.85\text{N}/\text{m}$, the spring is compressed by $8.00\text{cm}$. The mass of left and right side block is $0.250\text{kg}$ and $0.500\text{kg}$ respectively.

Write the expression to calculate the force by the spring.

    $F_{\text{s}}=kx$        (1)

Here, $F_{\text{s}}$ is the spring force, $k$ is spring constant and $x$ is the compression or expansion length.

Substitute $3.85\text{N}/\text{m}$ for $k$ and $8.00\text{cm}$ for $x$ in equation (1) to find $F_{\text{s}}$.

    $\begin{array}{c}{F}_{\text{s}}=3.85\text{N}/\text{m}\times 8.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\\ =0.308\text{N}\end{array}$

Write the expression of conservation of energy.

    $\begin{array}{c}{E}_{\text{spring}}=K_{\text{LB}}+K_{\text{RB}}\\ \frac{1}{2}k{x}^2=\frac{1}{2}{m}_1{v}_{\text{1f}}^{\text{2}}+\frac{1}{2}{m}_1{v}_{\text{2f}}^{\text{2}}\end{array}$        (2)

Here, $m_1$ is the mass of left block, $m_2$ is the mass of right block, $v_{1\text{f}}$ is the velocity of the left block and $v_{\text{2f}}$ is the velocity of the right block.

Substitute $3.85\text{N}/\text{m}$ for $k$, $8.00\text{cm}$ for $x$, $0.250\text{kg}$ for $m_1$ and $0.500\text{kg}$ for $m_2$ in equation (2).

    $\begin{array}{c}\frac{1}{2}\left(3.85\text{N}/\text{m}\right)\times {\left(8.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2=\frac{1}{2}\left(0.250\text{kg}\right)v_{\text{1f}}^{\text{2}}+\frac{1}{2}\left(0.500\text{kg}\right)v_{\text{2f}}^{\text{2}}\\ 0.0123\text{J}=\frac{1}{2}\left(0.250\text{kg}\right)v_{\text{1f}}^{\text{2}}+\frac{1}{2}\left(0.500\text{kg}\right)v_{\text{2f}}^{\text{2}}\end{array}$        (3)

Write the expression of conservation of linear momentum.

    $0=m_1{v}_{1\text{f}}+\left(m_2{v}_{\text{2f}}\right)$        (4)

Substitute $0.250\text{kg}$ for $m_1$ and $0.500\text{kg}$ for $m_2$ in equation (4).

    $\begin{array}{c}0=0.250\text{kg}\times v_{1\text{f}}+\left(0.500\text{kg}\times v_{\text{2f}}\right)\\ v_{1\text{f}}=-2v_{\text{2f}}\end{array}$        (5)

Substitute $2v_{\text{2f}}$ for $v_{1\text{f}}$ in equation (3) to find $v_{\text{2f}}$.

    $\begin{array}{c}0.0123\text{J}=\frac{1}{2}\left(0.250\text{kg}\right){\left(-2v_{\text{2f}}\right)}^2+\frac{1}{2}\left(0.500\text{kg}\right)v_{\text{2f}}^{\text{2}}\\ 0.0123\text{J}=\frac{1}{2}\left(1.50\text{kg}\right)v_{\text{2f}}^{\text{2}}\\ v_{\text{2f}}=0.128\widehat{\text{i}}\text{m}/\text{s}\end{array}$

Conclusion:

Substitute $0.128\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{2f}}$ in equation (5) to find $v_{1\text{f}}$.

    $\begin{array}{c}{v}_{1\text{f}}=-2\left(0.128\widehat{\text{i}}\text{m}/\text{s}\right)\\ =-0.256\widehat{\text{i}}\text{m}/\text{s}\end{array}$

Thus, the maximum velocity of each block is $-0.256\widehat{\text{i}}\text{m}/\text{s}$ and $0.128\widehat{\text{i}}\text{m}/\text{s}$ respectively.

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is $0.100$.

Answer to Problem 9.87AP

The maximum velocity of each block is $-0.0642\widehat{\text{i}}\text{m}/\text{s}$ and $0$ respectively.

Explanation of Solution

Write the expression to calculate the normal force on the lighter block.

    $R=m_{1}a$        (6)

Here, $a$ is the acceleration of the left block.

Substitute $0.250\text{kg}$ for $m_1$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (6) to find $R$.

    $\begin{array}{c}R=0.250\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =2.45\text{N}\end{array}$

Write the expression to calculate the limiting frictional force.

    $f_{\text{k}}=\mu R$

Here, $\mu$ is the coefficient of kinetic friction.

Substitute $2.45\text{N}$ for $R$ and $0.100$ for $\mu$ in above equation.

    $\begin{array}{c}{f}_{\text{k}}=0.100\times 2.45\text{N}\\ =\text{0}.245\text{N}\end{array}$

The spring force, $0.308\text{N}$ is greater than the limiting frictional force $\text{0}.245\text{N}$. Therefore, the left block will move.

Since the mass of right block is double than the left block therefore the limiting force of friction is also two times the left one i.e. $2f_{\text{k}}=0.490\text{N}$.

The limiting frictional force of right clock is greater than the spring force so it will not move.

The left will continue to move as long as the spring force is large than the friction force.

Write the expression to calculate the limiting frictional force.

    $f_{\text{k}}=k{x}_{\text{f}}$

Here, $x_{\text{f}}$ is the maximum distance travelled by the left block.

Substitute $\text{0}.245\text{N}$ for $f_{\text{k}}$ and $3.85\text{N}/\text{m}$ for $k$ in above equation.

    $\begin{array}{c}\text{0}.245\text{N}=3.85\text{N}/\text{m}\times x_{\text{f}}\\ x_{\text{f}}=0.0636\text{m}\end{array}$

Write the expression of conservation of energy.

    $\frac{1}{2}k{x}^2-f_{\text{k}}\left(x-x_{\text{f}}\right)=\frac{1}{2}{m}_1{v}_{\text{1f}}^{\text{2}}+\frac{1}{2}k{x}_{\text{f}}^{\text{2}}$        (7)

Substitute $3.85\text{N}/\text{m}$ for $k$, $8.00\text{cm}$ for $x$, $0.250\text{kg}$ for $m_1$, $\text{0}.245\text{N}$ for $f_{\text{k}}$, $0.0636\text{m}$ for $x_{\text{f}}$ in equation (7) to find $v_{1\text{f}}$.

  $\begin{array}{c}\left[\begin{array}{l}\frac{1}{2}\left(3.85\text{N}/\text{m}\right)\times {\left(8.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ -\left(\text{0}.245\text{N}\right)\left(8.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}-0.0636\text{m}\right)\end{array}\right]=\left[\begin{array}{l}\frac{1}{2}\left(0.250\text{kg}\right)v_{\text{1f}}^{\text{2}}\\ +\frac{1}{2}\left(3.85\text{N}/\text{m}\right){\left(0.0636\text{m}\right)}^2\end{array}\right]\\ 0.0123\text{J-0}\text{.004}\text{J}=\frac{1}{2}\left(0.250\text{kg}\right)v_{\text{1f}}^{\text{2}}+0.0078\text{J}\\ {\text{v}}_{1\text{f}}=0.0642\text{m}/\text{s}\\ =-0.0642\widehat{\text{i}}\text{m}/\text{s}\end{array}$

The negative sign indicates that the direction of motion of the lighter block is toward negative x axis.

The velocity of the heavier block is zero.

Conclusion:

Therefore, the maximum velocity of each block is $-0.0642\widehat{\text{i}}\text{m}/\text{s}$ and $0$ respectively.

To determine

The maximum velocity of each block if the coefficient of kinetic friction between block and the surface is $0.462$.

Answer to Problem 9.87AP

The velocity of both blocks is 0.

Explanation of Solution

Write the expression to calculate the limiting frictional force of left block.

    $\begin{array}{c}{f}_{\text{k}}=\mu R\\ =\mu m_{1}a\end{array}$

Substitute $0.250\text{kg}$ for $m_1$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.462$ for $\mu$ in above equation.

    $\begin{array}{c}{f}_{\text{k}}=0.462\times 0.250\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =1.13\text{N}\end{array}$

Write the expression to calculate the limiting frictional force of right block.

    $\begin{array}{c}{f}_{\text{k}}=\mu R\\ =\mu m_{2}a\end{array}$

Substitute $0.500\text{kg}$ for $m_2$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.462$ for $\mu$ in above equation.

    $\begin{array}{c}{f}_{\text{k}}=0.462\times 0.500\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =2.26\text{N}\end{array}$

The spring force is less than the limiting frictional force of both the blocks so the blocks will not move.

Conclusion:

Therefore, the velocity of both blocks is 0.