Chapter 13.1, Problem 13.30P

A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.

Fig. P13.30

To determine

Find the velocity (v) of the block after it has moved down $50\text{mm}$.

Answer to Problem 13.30P

The velocity (v) of the block after it has moved down $50\text{mm}$ is $\underset{\_}{0.597\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The mass of the block (m) is $10\text{kg}$.

The spring constant at A $\left(k_A\right)$ is $2\text{kN}/\text{m}$.

The spring constant at B $\left(k_B\right)$ is $2\text{kN}/\text{m}$.

The depth where the spring A moves down $\left(x_A\right)$ is $50\text{mm}$ or $0.05\text{m}$.

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

Calculation:

Show the free body diagram of the block with two spring’s attachment acting as in Figure (1).

Calculate the depth of spring B moves down due to block $\left(x_B\right)$ using the formula:

$x_B=\frac{1}{2}{x}_A$

Substitute $0.05\text{m}$ for $x_A$.

$\begin{array}{c}{x}_B=\frac{1}{2}\left(0.05\right)\\ =0.025\text{m}\end{array}$

Calculate the weight of the block (W) using the relation:

$W=mg$

Substitute $10\text{kg}$ for m and $9.81\text{m}/{\text{s}}^2$ for g.

$\begin{array}{c}W=\left(10\right)\left(9.81\right)\\ =98.1\text{N}\end{array}$

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

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

$T_2=\frac{1}{2}m{v}^2$

Substitute $10\text{kg}$ for m.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(10\right)v^2\\ =5v^2\end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_G$ due to gravity using the formula:

${\left(U_{1-2}\right)}_G=W{x}_A$

Substitute $98.1\text{N}$ for W and $0.05\text{m}$ for $x_A$.

$\begin{array}{c}{\left(U_{1-2}\right)}_G=\left(98.1\right)0.05\\ =4.905\text{N}\cdot \text{m}\end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_A$ due to spring A using the formula:

${\left(U_{1-2}\right)}_A=\frac{1}{2}{k}_A{x}_A^2$

Substitute $0.05\text{m}$ for $x_A$ and $2\text{kN}/\text{m}$ for $k_A$.

$\begin{array}{c}{\left(U_{1-2}\right)}_A=\frac{1}{2}\left(2\times {10}^3\right){\left(0.05\right)}^2\\ =2.5\text{N}\cdot \text{m}\end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_B$ due to spring B using the formula:

${\left(U_{1-2}\right)}_B=\frac{1}{2}{k}_B{x}_B^2$

Substitute $0.025\text{m}$ for $x_B$ and $2\text{kN}/\text{m}$ for $k_B$.

$\begin{array}{c}{\left(U_{1-2}\right)}_B=\frac{1}{2}\left(2\times {10}^3\right){\left(0.025\right)}^2\\ =0.625\text{N}\cdot \text{m}\end{array}$

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

$\left(U_{1-2}\right)={\left(U_{1-2}\right)}_G-{\left(U_{1-2}\right)}_A-{\left(U_{1-2}\right)}_B$

Substitute $4.905\text{N}\cdot \text{m}$ for ${\left(U_{1-2}\right)}_G$, $2.5\text{N}\cdot \text{m}$ for ${\left(U_{1-2}\right)}_A$, and $\text{0}\text{.625N}\cdot \text{m}$ for ${\left(U_{1-2}\right)}_B$.

$\begin{array}{c}\left(U_{1-2}\right)=4.905-2.5-0.625\\ =1.78\text{N}\cdot \text{m}\end{array}$

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.

Find the velocity (v) of the block after it has moved down $50\text{mm}$:

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

Substitute 0 for $T_1$, $1.78\text{N}\cdot \text{m}$ for $U_{1-2}$, and $5v^2$ for $T_2$.

$\begin{array}{l}0+\left(\text{1}\text{.78N}\cdot \text{m}\right)=5v^2\\ v=\sqrt{\frac{1.78}{5}}\\ v=0.597\text{m}/\text{s}\end{array}$

Therefore, the velocity (v) of the block after it has moved down $50\text{mm}$ is $\underset{\_}{0.597\text{m}/\text{s}}$.

To determine

Find the maximum velocity $\left(v_{\max }\right)$ achieved by the block.

Answer to Problem 13.30P

The maximum velocity $\left(v_{\max }\right)$ achieved by the block is $\underset{\_}{0.620\text{m}/\text{s}}$

Explanation of Solution

Given information:

The mass of the block (m) is $10\text{kg}$.

The spring constant at A $\left(k_A\right)$ is $2\text{kN}/\text{m}$.

The spring constant at B $\left(k_B\right)$ is $2\text{kN}/\text{m}$.

The depth where the spring A moves down $\left(x_A\right)$ is $50\text{mm}$ or $0.05\text{m}$.

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

Calculation:

Assume x be the distance moved down by the $10\text{kg}$ block.

Calculate the work done ${\left(U_{1-2}\right)}_G$ due to gravity using the formula:

${\left(U_{1-2}\right)}_G=Wx$

Substitute $98.1\text{N}$ for W.

${\left(U_{1-2}\right)}_G=98.1x$

Calculate the work done ${\left(U_{1-2}\right)}_A$ due to spring A using the formula:

${\left(U_{1-2}\right)}_A=\frac{1}{2}{k}_A{x}^2$

Substitute $2\text{kN}/\text{m}$ for $k_A$.

$\begin{array}{c}{\left(U_{1-2}\right)}_A=\frac{1}{2}\left(2\times {10}^3\right)x^2\\ =1000x^2\end{array}$

Calculate the work done ${\left(U_{1-2}\right)}_B$ due to spring B using the formula:

${\left(U_{1-2}\right)}_B=\frac{1}{2}{k}_B{\left(\frac{x}{2}\right)}^2$

Substitute $2\text{kN}/\text{m}$ for $k_B$.

$\begin{array}{c}{\left(U_{1-2}\right)}_B=\frac{1}{2}\left(2\times {10}^3\right){\left(\frac{x}{2}\right)}^2\\ \text{=1000}\left(\frac{x^2}{4}\right)\\ =250x^2\end{array}$

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

$\left(U_{1-2}\right)={\left(U_{1-2}\right)}_G-{\left(U_{1-2}\right)}_A-{\left(U_{1-2}\right)}_B$

Substitute $98.1x$ for ${\left(U_{1-2}\right)}_G$, $1000x^2$ for ${\left(U_{1-2}\right)}_A$, and $250x^2$ for ${\left(U_{1-2}\right)}_B$.

$\begin{array}{c}\left(U_{1-2}\right)=98.1x-1000x^2-250x^2\\ =98.1x-1250x^2\end{array}$ (1)

Differentiate the above equation with respect to ‘x’.

$\frac{d{U}_{1-2}}{dx}=0$

Substitute $98.1x-1250x^2$ for $U_{1-2}$.

$\begin{array}{l}\frac{d}{dx}\left(98.1x-1250x^2\right)=0\\ 98.1-2500x=0\\ x=\frac{98.1}{2,500}\\ x=0.03924\text{m}\left(39.24\text{mm}\right)\end{array}$

Substitute $0.03924\text{m}$ for x in Equation (1).

$\begin{array}{c}\left(U_{1-2}\right)=98.1\left(0.03924\right)-1000{\left(0.03924\right)}^2-250{\left(0.03924\right)}^2\\ =1.925\text{N}\cdot \text{m}\end{array}$

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.

Find the maximum velocity $\left(v_{\max }\right)$ achieved by the block:

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

Substitute 0 for $T_1$, $1.925\text{N}\cdot \text{m}$ for $U_{1-2}$, and $5v^2$ for $T_2$.

$\begin{array}{l}0+\left(1.925\text{N}\cdot \text{m}\right)=5v^2\\ v=\sqrt{\frac{1.925}{5}}\\ v=0.620\text{m}/\text{s}\end{array}$

Therefore, the maximum velocity $\left(v_{\max }\right)$ achieved by the block is $\underset{\_}{0.620\text{m}/\text{s}}$.