Chapter 9, Problem 47PQ

(a)

To determine

The speed of the car when it reaches the bottom of the hill.

Answer to Problem 47PQ

The speed of the car when it reaches the bottom of the hill is $39.8\text{ m/s}$ .

Explanation of Solution

It is given that the rolling friction is negligible. This implies the principle of conservation of energy can be applied on the system. The gravitational potential energy of the car is converted to its kinetic energy as it falls to the bottom of the hill.

Write the expression for the conservation of energy.

  $K_i+U_i=K_f+U_i$                                                                                                  (I)

Here, $K_i$ is the initial kinetic energy of the car, $U_i$ is the initial gravitational potential energy of the car, $K_f$ is the final kinetic energy and $U_{gf}$ is the gravitational potential energy of the car at the bottom of the hill.

The initial kinetic energy of the car is zero.

Write the equation for $K_i$ .

  $K_i=0$                                                                                                                    (II)

Assume that initially the car is at a height $y$ .

Write the expression for $U_i$ .

  $U_i=mgy$                                                                                                             (III)

Here, $m$ is the mass of the car, $g$ is the acceleration due to gravity and $y$ is the initial height of the car.

Assume the height of the car at the bottom of the hill is zero.

Write the expression for $U_f$ .

  $U_f=0$                                                                                                     (IV)

Write the expression for $K_f$ .

  $K_f=\frac{1}{2}m{v}_f^2$                                                                                                          (V)

Here, $v_f$ is the speed of the car at the bottom of the hill.

Put equations (II) to (V) in equation (I) and rewrite the equation for $v_f$ .

  $\begin{array}{c}0+mgy=\frac{1}{2}m{v}_f^2+0\\ gy=\frac{1}{2}{v}_f^2\\ v_f^2=2gy\\ v_f=\sqrt{2gy}\end{array}$                                                                                  (VI)

Conclusion:

Given that the initial height of the car is $80.7\text{ m}$ . The value of acceleration due to gravity is $9.80{\text{ m/s}}^2$ .

Substitute $9.80{\text{ m/s}}^2$ for $g$ and $80.7\text{ m}$ for $y$ in equation (VI) to find $v_f$ .

  $\begin{array}{c}{v}_f=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(80.7\text{ m}\right)}\\ =39.8\text{ m/s}\end{array}$

Therefore, the speed of the car when it reaches the bottom of the hill is $39.8\text{ m/s}$ .

(b)

To determine

The amount of thermal energy of the system that changes during the stopping motion of the car.

Answer to Problem 47PQ

The amount of thermal energy of the system that changes during the stopping motion of the car is $5.94\times {10}^5\text{ J}$ .

Explanation of Solution

Write the expression for the work-energy theorem.

  $K_i+U_i+W_{\text{tot}}=K_f+U_f+\Delta E_{\text{th}}$                                                                           (VII)

Here, $W_{\text{tot}}$ is the total work done and $\Delta E_{\text{th}}$ is the change in thermal energy.

Define the system as the plastic track and the roller coaster’s wheels. During the last stretch on the track, the coaster will be at the same height as that of its initial height. This implies there is no change in potential energy of the system.

  $U_f=U_i$

There are no external forces doing work on the system.

Write the expression for $W_{\text{tot}}$ .

  $W_{\text{tot}}=0$

The roller coaster is stopped at the final configuration so that its final kinetic energy will be zero.

Write the expression for $K_f$ .

  $K_f=0$

Put the above three equations in equation (VII) .

  $\begin{array}{c}{K}_i+U_i+0=0+U_i+\Delta E_{\text{th}}\\ \Delta E_{\text{th}}=K_i\end{array}$                                                                                 (VIII)

Write the expression for $K_i$ .

  $K_i=\frac{1}{2}m{v}_i^2$                                                                                                      (IX)

Here, $v_i$ is the speed of the car when it reaches the bottom of the hill.

Conclusion:

In part (a), it is found that the value of $v_i$ is $39.8\text{ m/s}$ . It is given that the mass of the roller-coaster car is $750.0\text{ kg}$ .

Substitute $750.0\text{ kg}$ for $m$ and $39.8\text{ m/s}$ for $v_i$ in equation (IX) to find $K_i$ .

  $\begin{array}{c}{K}_i=\frac{1}{2}\left(750.0\text{ kg}\right){\left(39.8\text{ m/s}\right)}^2\\ =5.94\times {10}^5\text{ J}\end{array}$

Substitute $5.94\times {10}^5\text{ J}$ for $K_i$ in equation (VIII) to find $\Delta E_{\text{th}}$ .

  $\Delta E_{\text{th}}=5.94\times {10}^5\text{ J}$

Therefore, the amount of thermal energy of the system that changes during the stopping motion of the car is $5.94\times {10}^5\text{ J}$ .

(c)

To determine

The coefficient of kinetic friction between the wheels and the plastic stopping track.

Answer to Problem 47PQ

The coefficient of kinetic friction between the wheels and the plastic stopping track is $0.350$ .

Explanation of Solution

The change in thermal energy of the system occurs due to the work done by the frictional force on the system.

Write the expression for $\Delta E_{\text{th}}$ .

  $\Delta E_{\text{th}}=W_k$                                                                                                         (X)

Here, $W_k$ is the work done by the kinetic frictional force on the system.

Write the expression for $W_k$ .

$W_k=F_{k}s$                                                                                                        (XI)

Here, $F_k$ is the force of kinetic friction and $s$ is the distance travelled by the car before it stops.

Write the expression for $F_k$ .

  $F_k={\mu }_k{F}_N$                                                                                                             (XII)

Here, ${\mu }_k$ is the coefficient of kinetic friction and $F_N$ is the normal force.

Write the expression for $F_N$ .

  $F_N=mg$

Put the above equation in equation (XII).

  $F_k={\mu }_{k}mg$

Put the above equation in equation (XI).

  $W_k={\mu }_{k}mgs$

Put the above equation in equation (X) and rewrite it for ${\mu }_k$ .

  $\begin{array}{c}\Delta E_{\text{th}}={\mu }_{k}mgs\\ {\mu }_k=\frac{\Delta E_{\text{th}}}{mgs}\end{array}$                                                                                                    (XIII)

Conclusion:

It is given that the car stops at $230.66\text{ m}$ .

Substitute $5.94\times {10}^5\text{ J}$ for $\Delta E_{\text{th}}$, $750.0\text{ kg}$ for $m$ , $9.80{\text{ m/s}}^2$ for $g$ and $230.66\text{ m}$ for $s$ in equation (XIII) to find ${\mu }_k$ .

  $\begin{array}{c}{\mu }_k=\frac{5.94\times {10}^5\text{ J}}{\left(750.0\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(230.66\text{ m}\right)}\\ =0.350\end{array}$

Therefore, the coefficient of kinetic friction between the wheels and the plastic stopping track is $0.350$ .