Chapter 8, Problem 25AP

(a)

To determine

The constant friction force for the last $20\text{m}$ for the empty test car.

Answer to Problem 25AP

The constant friction force for the last $20\text{m}$ for the empty test car is $1.29\times {10}^4\text{N}$.

Explanation of Solution

Given info: The mass of the empty car is $250\text{kg}$, the height of the track is $110\text{m}$, the total length of the ground is $250\text{m}$.

The formula to calculate the energy by equation of motion is,

$K_{\text{i}}+U_{\text{i}}+\Delta E=K_{\text{f}}+U_{\text{f}}$ (1)

Here,

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

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

$\Delta E$ is the energy wasted.

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

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

The formula of kinetic energy at start is,

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

Substitute $0$ for $v_{\text{i}}$ in the above equation to find the value of $K_{\text{i}}$.

$\begin{array}{c}{K}_{\text{i}}=\frac{1}{2}m\left(0^2\right)\\ =0\end{array}$

The formula of kinetic energy at final point is,

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

Substitute $0$ for $v_{\text{f}}$ in the above equation to find the value of $K_{\text{f}}$.

$\begin{array}{c}{K}_{\text{f}}=\frac{1}{2}m\left(0^2\right)\\ =0\end{array}$

The formula of the potential energy at start is,

$U_{\text{i}}=mgh$

Substitute $y_{\text{i}}$ for $h$ in the above equation to find the value of $U_{\text{i}}$.

$U_{\text{i}}=mg{y}_{\text{i}}$

The formula of the potential energy at final point is,

$U_{\text{f}}=mgh$

Substitute $0$ for $h$ in the above equation to find the value of $U_{\text{f}}$.

$U_{\text{f}}=0$

The formula of the energy wasted due to friction is,

$\Delta E=f\Delta x$

Substitute $-f_1\Delta x_1-f_2\Delta x_2$ for $f\Delta x$ in the above equation to find the value of $\Delta E$.

$\Delta E=-f_1\Delta x_1-f_2\Delta x_2$

Substitute $0$ for $K_{\text{i}}$, $mg{y}_{\text{i}}$ for $U_{\text{i}}$, $-f_1\Delta x_1-f_2\Delta x_2$ for $\Delta E$, $0$ for $U_{\text{f}}$ and $0$ for $K_{\text{f}}$ in the equation (1) to find the value of $f_2$ .

$0+mg{y}_{\text{i}}-f_1\Delta x_1-f_2\Delta x_2=0+0$

Substitute $250\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $110\text{m}$ for $y_{\text{i}}$, $50\text{N}$ for $f_1$, $230\text{m}$ for $\Delta x_1$ and $20\text{m}$ for $\Delta x_2$ in the above equation to find the value of $f_2$.

$\begin{array}{c}\left(250\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(110\text{m}\right)-\left(50\text{N}\right)\left(230\text{m}\right)-f_2\left(20\text{m}\right)=0\\ f_2=12900\text{N}\\ =1.29\times {10}^4\text{N}\end{array}$

Conclusion:

Therefore, the constant friction force for the last $20\text{m}$ for the empty test car is $1.29\times {10}^4\text{N}$.

(b)

To determine

The highest speed reached by the car.

Answer to Problem 25AP

The highest speed reached by the car is $45.5\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the empty car is $250\text{kg}$, the height of the track is $110\text{m}$, the total length of the ground is $250\text{m}$.

The formula of kinetic energy at final point is,

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

The body is at the lowest point so the resistive force is taken upto that point only.

The formula of the energy wasted due to friction is,

$\Delta E=f\Delta x$

Substitute $-f_1\Delta x_1$ for $f\Delta x$ in the above equation to find the value of $\Delta E$.

$\Delta E=-f_1\Delta x_1$

The formula to calculate the final velocity from equation (1) can be written as,

$mg{y}_{\text{i}}-f_1\Delta x_1=\frac{1}{2}m{v}_{\text{f}}{}^2$ (2)

Substitute $250\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $110\text{m}$ for $y_{\text{i}}$, $50\text{N}$ for $f_1$ and $230\text{m}$ for $\Delta x_1$ in equation (2) to find the value of $v_{\text{f}}$.

$\begin{array}{c}\left(250\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(110\text{m}\right)-\left(50\text{N}\right)\left(230\text{m}\right)=\frac{1}{2}\left(250\text{kg}\right)v_{\text{f}}{}^2\\ 125v_{\text{f}}{}^2=258000\\ v_{\text{f}}=45.5\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the highest speed reached by the car is $45.5\text{m}/\text{s}$.

(c)

To determine

The new values of friction force and highest speed when the weight is $450\text{kg}$ .

Answer to Problem 25AP

The new value of friction force is $3.72\times {10}^4\text{N}$ and highest speed when the weight is $450\text{kg}$ is $46.1\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the empty car is $250\text{kg}$, the height of the track is $110\text{m}$, the total length of the ground is $250\text{m}$.

The formula to calculate the force of friction form equation (1) is,

$0+mg{y}_{\text{i}}-f_1\Delta x_1-f_2\Delta x_2=0+0$

Substitute $450\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $110\text{m}$ for $y_{\text{i}}$, $50\text{N}$ for $f_1$, $230\text{m}$ for $\Delta x_1$ and $20\text{m}$ for $\Delta x_2$ in the above equation to find the value of $f_2$.

$\begin{array}{c}\left(450\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(110\text{m}\right)-\left(50\text{N}\right)\left(230\text{m}\right)-f_2\left(20\text{m}\right)=0\\ f_2=37200\text{N}\\ =3.72\times {10}^4\text{N}\end{array}$

To calculate the final velocity the equation (2) is,

$mg{y}_{\text{i}}-f_1\Delta x_1=\frac{1}{2}m{v}_{\text{f}}{}^2$

Substitute $450\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $110\text{m}$ for $y_{\text{i}}$, $50\text{N}$ for $f_1$ and $230\text{m}$ for $\Delta x_1$ in the above equation to find the value of $v_{\text{f}}$.

$\begin{array}{c}\left(450\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(110\text{m}\right)-\left(50\text{N}\right){\left(230\text{m}\right)}_1=\frac{1}{2}\left(450\text{kg}\right)v_{\text{f}}{}^2\\ 225v_{\text{f}}{}^2=473600\\ v_{\text{f}}=46.1\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the new value of friction force is $3.72\times {10}^4\text{N}$ and highest speed when the weight is $450\text{kg}$ is $46.1\text{m}/\text{s}$.

(d)

To determine

The depth of the underground part of the ride.

Answer to Problem 25AP

The depth of the underground part of the ride is $45\text{m}$.

Explanation of Solution

Consider $x$ be the depth of the underground tunnel made beyond height $y_{\text{i}}$

The formula of the potential energy at start is,

$U_{\text{i}}=mgh$

Substitute $\left(y_{\text{i}}+x\right)$ for $h$ in the above equation to find the value of $U_{\text{i}}$.

$U_{\text{i}}=mg\left(y_{\text{i}}+x\right)$

The formula to calculate the depth is,

$mg\left(y_{\text{i}}+x\right)-f_1\Delta x_1=\frac{1}{2}m{v}_{\text{f}}{}^2$

Substitute $450\text{kg}$ for $m$, $9.85\text{m}/{\text{s}}^{\text{2}}$ for $g$, $110\text{m}$ for $y_{\text{i}}$, $50\text{N}$ for $f_1$, $55\text{m}/{\text{s}}^{\text{2}}$ for $v_{\text{f}}$ and $150\text{m}$ for $\Delta x_1$ in the above equation to find the value of $x$.

$\begin{array}{l}\left(450\text{kg}\right)\left(9.85\text{m}/{\text{s}}^{\text{2}}\right)\left(\left(110\text{m}\right)+x\right)-\left(50\text{N}\right)\left(150\text{m}\right)=\frac{1}{2}\left(450\text{kg}\right){\left(55\text{m}/{\text{s}}^{\text{2}}\right)}^2\\ 4432.5\left(110+x\right)=688125\\ 110+x=155.2\\ x=45.2\approx 45\text{m}\end{array}$

$x\approx 45\text{m}$

Conclusion:

Therefore, the depth of the underground part of the ride is $45\text{m}$.

(d)

To determine

The depth of the underground provided is feasible or not.

Answer to Problem 25AP

No, the depth of the underground provided is not feasible.

Explanation of Solution

No the depth of the underground tunnel provided is not feasible as the total length covered by the car is increased due to which the loss of energy due to friction is increased a lot as compared to the case when the underground path was not there. The non conventional forces on the car are increased due to the depth provided as the friction in the path of later $20\text{m}$ is very high. So, the depth of the underground provided is not feasible.

Conclusion:

Therefore, the depth of the underground provided is not feasible.