Chapter 13.1, Problem 13.26P

A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

Fig. P13.26

To determine

Find the maximum speed $\left(v_2\right)$ reached by the $2\text{kg}$ block.

Answer to Problem 13.26P

The maximum speed $\left(v_2\right)$ reached by the $2\text{kg}$ block is $\underset{\_}{3.290\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The mass of the block A $\left(m_A\right)$ is $2\text{kg}$.

The mass of the block B $\left(m_B\right)$ is $3\text{kg}$.

The spring constant (k) is $40\text{N}/\text{m}$.

Assume the acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Consider the position 1, the block B has been removed.

Calculate the spring stretch $\left(x_1\right)$ using the relation:

$F_S=-k{x}_1$

Here, $F_S$ is the spring force.

Substitute $-\left(m_A+m_B\right)g$ for $F_S$.

$\begin{array}{l}\left(m_A+m_B\right)g=-k{x}_1\\ x_1=\frac{-\left(m_A+m_B\right)g}{k}\end{array}$

Substitute $2\text{kg}$ for $m_A$, $3\text{kg}$ for $m_B$, $9.81\text{m}/{\text{s}}^2$ for g, and $40\text{N}/\text{m}$ for k.

$\begin{array}{c}{x}_1=\frac{-\left(2+3\right)9.81}{40}\\ =-1.22625\text{m}\end{array}$

Take the position 2 be later position while the spring still in contact with block A.

Calculate the work of the force exerted ${\left(U_{1-2}\right)}_e$ by the spring using the formula:

${\left(U_{1-2}\right)}_e=-{\displaystyle \underset{x_1}{\overset{x_2}{\int }}kxdx}$

Integrate the above equation with respect to ‘x’.

$\begin{array}{c}{\left(U_{1-2}\right)}_e=-{\left[\frac{k{x}^2}{2}\right]}_{x_1}^{x_2}\\ =-\frac{k{x}_2^2}{2}-\left(-\frac{k{x}_1^2}{2}\right)\\ =\frac{k{x}_1^2}{2}-\frac{k{x}_2^2}{2}\end{array}$

Substitute, $-1.22625\text{m}$ for $x_1$ and $40\text{N}/\text{m}$ for k.

$\begin{array}{c}{\left(U_{1-2}\right)}_e=\frac{40{\left(-1.22625\right)}^2}{2}-\frac{40x_2^2}{2}\\ =30.074-20x_2^2\end{array}$

Calculate the work of the gravitational force ${\left(U_{1-2}\right)}_g$ using the formula:

${\left(U_{1-2}\right)}_g=-m_{A}g\left(x_2-x_1\right)$

Substitute, $-1.22625\text{m}$ for $x_1$, $9.81\text{m}/{\text{s}}^2$ for g, and $2\text{kg}$ for $m_A$.

$\begin{array}{c}{\left(U_{1-2}\right)}_g=-2\times 9.81\left(x_2+1.22625\right)\\ =-19.62x_2-24.059\end{array}$

Calculate the total work done $\left(U_{1-2}\right)$ using the formula:

$\left(U_{1-2}\right)={\left(U_{1-2}\right)}_e+{\left(U_{1-2}\right)}_g$

Substitute $30.074-20x_2^2$ for ${\left(U_{1-2}\right)}_e$ and $-19.62x_2-24.059$ for ${\left(U_{1-2}\right)}_g$.

$\begin{array}{c}\left(U_{1-2}\right)=30.074-20x_2^2+\left(-19.62x_2-24.059\right)\\ =-20x_2^2-19.62x_2+6.015\end{array}$

Here, the initial kinetic energy $\left(T_1\right)$ is zero.

Calculate the kinetic energy $\left(T_2\right)$ using the formula:

$\left(T_2\right)=\frac{1}{2}{m}_A{v}_A^2$

Substitute $2\text{kg}$ for $m_A$.

$\begin{array}{c}\left(T_2\right)=\frac{1}{2}\left(2\right)v_2^2\\ =v_2^2\end{array}$

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Write the expression for the principle of work and energy:

$T_1+U_{1-2}=T_2$

Substitute 0 for $T_1$, $-20x_2^2-19.62x_2+6.015$ for $U_{1-2}$, and $v_2^2$ for $T_2$.

$\begin{array}{l}0+\left(-20x_2^2-19.62x_2+6.015\right)=v_2^2\\ v_2^2=-20x_2^2-19.62x_2+6.015\end{array}$ (1)

At the maximum speed, differentiate the velocity equation with respect to ‘x’.

$\frac{d{v}_2}{d{x}_2}=0$

Substitute $-20x_2^2-19.62x_2+6.015$ for $v_2$.

$\begin{array}{l}\frac{d}{d{x}_2}\left(-20x_2^2-19.62x_2+6.015\right)=0\\ -40x_2-19.62\left(1\right)+6.015\left(0\right)=0\\ -40x_2-19.62=0\end{array}$

$\begin{array}{l}-40x_2=19.62\\ x_2=-0.4905\text{m}\end{array}$

Substitute $-0.4905\text{m}$ for $x_2$.

$\begin{array}{c}{v}_2^2=-20{\left(-0.4905\right)}^2-19.62\left(-0.4905\right)+6.015\\ v_2^2=10.82681\\ v_2=3.290\text{m}/\text{s}\end{array}$

Therefore, the maximum speed $\left(v_2\right)$ reached by the $2\text{kg}$ block is $\underset{\_}{3.290\text{m}/\text{s}}$.

To determine

Find the maximum height (h) reached by the $2\text{kg}$ block.

Answer to Problem 13.26P

The maximum height (h) reached by the $2\text{kg}$ block is $\underset{\_}{1.533\text{m}}$.

Explanation of Solution

Given information:

The mass of the block A $\left(m_A\right)$ is $2\text{kg}$.

The mass of the block B $\left(m_B\right)$ is $3\text{kg}$.

The spring constant (k) is $40\text{N}/\text{m}$.

Assume the acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Consider the position 3, the block A reached the maximum height and assume that the block has separated from the spring so the spring is zero at the separation.

Calculate the work of the force exerted ${\left(U_{1-3}\right)}_e$ by the spring using the formula:

${\left(U_{1-3}\right)}_e=-{\displaystyle \underset{x_1}{\overset{0}{\int }}kxdx}$

Integrate the above equation with respect to ‘x’.

$\begin{array}{c}{\left(U_{1-3}\right)}_e=-{\left[\frac{k{x}^2}{2}\right]}_{x_1}^0\\ =-\frac{k{\left(0\right)}^2}{2}-\left(-\frac{k{x}_1^2}{2}\right)\\ =\frac{k{x}_1^2}{2}\end{array}$

Substitute, $-1.22625\text{m}$ for $x_1$ and $40\text{N}/\text{m}$ for k.

$\begin{array}{c}{\left(U_{1-3}\right)}_e=\frac{40{\left(-1.22625\right)}^2}{2}\\ =30.074\text{J}\end{array}$

Calculate the work of the gravitational force ${\left(U_{1-3}\right)}_g$ using the formula:

${\left(U_{1-3}\right)}_g=-m_{A}gh$

Substitute $9.81\text{m}/{\text{s}}^2$ for g and $2\text{kg}$ for $m_A$.

$\begin{array}{c}{\left(U_{1-3}\right)}_g=-2\times 9.81\times h\\ =-19.62h\end{array}$

Calculate the total work done $\left(U_{1-3}\right)$ using the formula:

$\left(U_{1-3}\right)={\left(U_{1-3}\right)}_e+{\left(U_{1-3}\right)}_g$

Substitute $30.074\text{J}$ for ${\left(U_{1-3}\right)}_e$ and $-19.62h$ for ${\left(U_{1-3}\right)}_g$.

$\begin{array}{c}\left(U_{1-3}\right)=30.074\text{J}+\left(-19.62h\right)\\ =30.075-19.62h\end{array}$

At the maximum height, the velocity $\left(v_3\right)$ and kinetic energy $\left(T_3\right)$ is zero respectively.

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Find the maximum height (h) reached by the $2\text{kg}$ block principle of work and energy:

$T_1+U_{1-3}=T_3$

Substitute 0 for $T_1$, $30.075-19.62h$ for $U_{1-3}$, and 0 for $T_3$.

$\begin{array}{l}0+30.075-19.62h=0\\ 19.62h=30.075\\ h=1.533\text{m}\end{array}$

Therefore, the maximum height (h) reached by the $2\text{kg}$ block is $\underset{\_}{1.533\text{m}}$.