Chapter 8, Problem 8.62AP

A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has a speed of v_i = 3.00 m/s when it makes contact with a light spring (Fig. P8.62b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Fig. P8.62c). The object is then forced toward the left by the spring (Fig. P8.62d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Fig. P8.62e). Find (a) the distance of compression d, (b) the speed vat the unstretched posi-tion when the object is moving to the left (Fig. P8.624), and (c) the distance D where the abject comes to rest.

Figure P8.62

To determine

The distance of the compression.

Answer to Problem 8.62AP

The distance of the compression is $0.378\text{m}$.

Explanation of Solution

Given info: The mass of the object is $1.0\text{kg}$, value of coefficient of kinetic friction is $0.250$, speed of the object is $3.00\text{m}/\text{s}$.

The formula to calculate the change in energy is,

$\Delta E_1=-\mu mgd$

Here,

$m$ is the mass of the object.

$g$ is the acceleration due to gravity.

$\mu$ is the coefficient of kinetic friction.

$d$ is the distance of compression of the spring.

The formula to calculate the initial kinetic energy of the object is,

$K_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2$

Here,

$m$ is the mass of the object.

$v_{\text{i}}$ is the initial velocity of the object.

The formula to calculate the final kinetic energy is,

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2$

Here,

$m$ is the mass of the object.

$v_{\text{f}}$ is the final velocity of the object.

The formula to calculate initial potential energy is,

$U_{\text{i}}=\frac{1}{2}m{x}_{\text{i}}{}^2$

Here,

$m$ is the mass of the block.

$x_{\text{i}}$ is the initial compression distance.

Thus the initial potential energy of the block is $\frac{1}{2}m{x}_{\text{i}}{}^2$.

The formula to calculate the final potential energy is,

$U_{\text{f}}=\frac{1}{2}k{x}_{\text{f}}{}^2$

Here,

$k$ is the spring constant.

$x_{\text{f}}$ is the final distance of compression.

Thus, the final potential energy of the block is $\frac{1}{2}k{x}_{\text{f}}{}^2$.

The formula to calculate the initial energy is,

$E_{\text{i}}=U_{\text{i}}+K_{\text{i}}$

Here,

$U_{\text{i}}$ is the initial potential energy.

$K_{\text{i}}$ is the initial kinetic energy.

Substitute $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K_{\text{i}}$ and $\frac{1}{2}k{x}_{\text{i}}{}^2$ for $U_{\text{i}}$ in the above formula to find $E_{\text{i}}$.

$\begin{array}{c}{E}_{\text{i}}=U_{\text{i}}+K_{\text{f}}\\ =\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2\end{array}$

Thus, the initial energy is $\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2$.

The formula to calculate the final energy is,

$E_{\text{f}}=U_{\text{f}}+K_{\text{f}}$

Here,

$U_{\text{f}}$ is the final potential energy.

$K_{\text{f}}$ is the final kinetic energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_{\text{f}}$ and $\frac{1}{2}k{x}_{\text{f}}{}^2$ for $U_{\text{f}}$ in the above formula to find $E_{\text{f}}$.

$\begin{array}{c}{E}_{\text{f}}=U_{\text{f}}+K_{\text{f}}\\ =\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2\end{array}$

Thus, the final energy is $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$.

The formula to calculate the law of conservation of energy between the second and third diagram is,

$E_{\text{f}}-E_{\text{i}}=\Delta E_1$

Here,

$E_{\text{f}}$ is the final energy.

$E_{\text{i}}$ is the initial energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$ for $E_{\text{f}}$ , $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2$ for $E_{\text{i}}$, in the above formula to find $\Delta E$.

$\begin{array}{l}{E}_{\text{f}}-E_{\text{i}}=\Delta E_1\\ \frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2-\frac{1}{2}m{v}_{\text{i}}{}^2+mgh=-\mu mgd\end{array}$

Substitute $0$ for $\frac{1}{2}m{v}_{\text{f}}{}^2$, $0$ for $\frac{1}{2}k{x}_{\text{i}}{}^2$ and $d$ for $x_{\text{f}}$ in the above formula to find $d$ .

$\frac{1}{2}k{d}^2-\frac{1}{2}m{v}_1{}^2=-\mu mgd$

Substitute $1.00\text{}\text{kg}$ for $m$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $0.250$  for $\mu$, $50\text{N}/\text{m}$ for $k$  , $3.00\text{m}/\text{s}$ for $v_1$ in the above formula to find $d$.

$\begin{array}{c}-\left(0.250\right)\left(1.00\text{}\text{kg}\right)d=\frac{1}{2}\left(50\text{N}/\text{m}\right)d^2-\left(1.00\text{}\text{kg}\right)\left(3.00\text{m}/\text{s}\right)\\ =\left(25\text{N}/\text{m}\right)d^2+\left(2.45\text{N}\right)d-4.5\text{J}\\ d=\frac{-2.45\text{N}\pm \sqrt{{\left(2.45\text{N}\right)}^2-4\left(25.0\text{}\text{N}\right)\left(-4.5\text{J}\right)}}{2\left(25.0\text{N}/\text{m}\right)}\end{array}$

Further solve the above equation.

$\begin{array}{c}d=\frac{-2.45\pm 21.35\text{N}}{50.0\text{N}/\text{m}}\\ =0.378\text{m}\end{array}$

Conclusion:

Therefore, the distance of the compression is $0.378\text{}\text{m}$.

To determine

The speed at the unstretched position when the object is moving to the left.

Answer to Problem 8.62AP

The speed at the unstretched position when the object is moving to the left is $2.30\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the object is $1.0\text{kg}$, value of coefficient of kinetic friction is $0.250$, speed of the object is $3.00\text{m}/\text{s}$.

The formula to calculate the change in energy is,

$\Delta E_2=-\mu mg\left(2d\right)$

Here,

$m$ is the mass of the object.

$g$ is the acceleration due to gravity.

$\mu$ is the coefficient of kinetic friction.

$d$ is the distance of compression of the spring.

The formula to calculate the law of conservation of energy between the second and fourth diagram is,

$E_{\text{f}}-E_{\text{i}}=\Delta E_2$

Here,

$E_{\text{f}}$ is the final energy.

$E_{\text{i}}$ is the initial energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$ for $E_{\text{f}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2$ for $E_{\text{i}}$, $-\mu mgd\left(2d\right)$ for $\Delta E_2$, in the above formula to find $\mu$.

$\begin{array}{l}{E}_{\text{f}}-E_{\text{i}}=\Delta E_2\\ \frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2-\frac{1}{2}m{v}_2{}^2+mgh=-\mu mgd\left(2d\right)\end{array}$

Substitute $0$ for $\frac{1}{2}k{x}_{\text{i}}{}^2$ , 0 for $\frac{1}{2}k{x}_{\text{f}}{}^2$, for in the above formula to find $v_{\text{f}}$ .

$\begin{array}{c}\frac{1}{2}m{v}_{\text{f}}{}^2-\frac{1}{2}m{v}_{\text{i}}{}^2=-\mu mg\left(2d\right)\\ v^2=\sqrt{v_1{}^2-\frac{4\left(\mu mg\right)d}{m}}\\ =\sqrt{v_1{}^2-4\mu mgd}\end{array}$

Substitute $1.00\text{}\text{kg}$ for $m$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $0.250$ for $\mu$ , $3.00\text{m}/\text{s}$ for $v_{\text{i}}$ in the above formula to find $v$.

$\begin{array}{c}{v}_{\text{f}}=\sqrt{v_{\text{i}}{}^2-4\mu mgd}\\ =\sqrt{{\left(3.00\text{m}/\text{s}\right)}^2-4\left(0.250\right)\left(0.378\text{m}\right)}\\ =2.30\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed at the unstretched position when the object is moving to the left is $2.30\text{m}/\text{s}$.

To determine

The distance where the object comes to rest.

Answer to Problem 8.62AP

The distance where the object comes to rest is $1.08\text{}\text{m}$.

Explanation of Solution

Given info: The mass of the object is $1.0\text{kg}$ , value of coefficient of kinetic friction is $0.250$, speed of the object is $3.00\text{m}/\text{s}$.

The formula to calculate the change in energy is,

$\Delta E_3=-\mu mg\left(D+2d\right)$

Here,

$m$ is the mass of the object.

$g$ is the acceleration due to gravity.

$\mu$ is the coefficient of kinetic friction.

$d$ is the distance of compression of the spring.

Thus, the value of change in energy is $-\mu mg\left(D+2d\right)$.

The formula to calculate the law of conservation of energy between the second and fifth diagram is,

$E_{\text{f}}-E_{\text{i}}=\Delta E_3$

Here,

$E_{\text{f}}$ is the final energy.

$E_{\text{i}}$ is the initial energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$ for $E_{\text{f}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2$ for $E_{\text{i}}$, $-\mu mgd\left(D+2d\right)$ for $\Delta E_3$, in the above formula to find $D$.

$\begin{array}{l}{E}_{\text{f}}-E_{\text{i}}=\Delta E_3\\ \frac{1}{2}m{v}_{\text{I}}{}^2+\frac{1}{2}k{x}_{\text{i}}{}^2-\frac{1}{2}m{v}_{\text{f}}{}^2+mgh=-\mu mgd\left(D+2d\right)\end{array}$

Substitute $0$ for $\frac{1}{2}k{x}_{\text{i}}{}^2$ , 0 for $\frac{1}{2}k{x}_{\text{f}}{}^2$, 0 for $\frac{1}{2}m{v}_{\text{f}}{}^2$ in the above formula to find $D$ .

$\begin{array}{c}-\frac{1}{2}m{v}_{\text{i}}{}^2=-\mu mg\left(D+2d\right)\\ \mu mg\left(D+2d\right)=\frac{1}{2}m{v}^2\\ D=\frac{v_{\text{i}}{}^2}{\mu g}-2d\end{array}$

Substitute $1.00\text{}\text{kg}$ for $m$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $0.250$ for $\mu$ , $3.00\text{m}/\text{s}$ for $v_{{}_{\text{i}}}$ in the above formula to find $D$.

$\begin{array}{c}D=\frac{v_{\text{i}}{}^2}{\mu g}-2d\\ =\frac{{\left(3.00\text{m}/\text{s}\right)}^2}{\left(0.250\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}-2\left(0.378\right)\\ =1.08\text{}\text{m}\end{array}$

Conclusion:

Therefore, the distance where the object comes to rest is $1.08\text{}\text{m}$.