Chapter 9, Problem 50PQ

A small 0.65-kg box is launched from rest by a horizontal spring as shown in Figure P9.50. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. Kinetic friction between the box and the track is negligible on the hill, but the coefficient of kinetic friction between the box and the horizontal parts of track is 0.35. The spring has a spring constant of 34.5 N/m, and is compressed 30.0 cm with the box attached. The block remains on the track at all times.

a. What would you include in the system? Explain your choice.

b. Calculate d.

To determine

The items included in the system and explain the choice.

Answer to Problem 50PQ

The items included are the box and the tracks surface because the kinetic friction increases the thermal energies and included this thermal energy internal to the system.

Explanation of Solution

In order to keep all the thermal energy into the system, it is better to include both the box and the tracks surface so that it will increase the kinetic friction which leads to the increase in the thermal energies and including both keeps all of this thermal energy internal to the system.

If Earth and spring are the choices, to account for them in terms of changes in gravitational and elastic potential energy without letting anything outside the system to do work.

Figure 1 show the graph of the initial and final energies which will help to organize the energies needed to be taken into account.

Conclusion:

Therefore, the items included are box and the tracks surface because the kinetic friction increases the thermal energies and included this thermal energy internal to the system.

To determine

The value of $d$.

Answer to Problem 50PQ

The value of $d$ is $\underset{\_}{10\text{m}}$.

Explanation of Solution

The reference configuration for the spring is when it is relaxed, and for gravity it is when the box is at the bottom of the ramp. The box is initially at rest $\left(K_i=0\right)$, there are no external forces $\left(W_{\text{total}}=0\right)$, and in the final situation, the box is again at rest at the reference height and the spring is relaxed $\left(K_f=0,U_{sf}=0,U_{gf}=0\right)$.

The energy conservation equation for a system is,

  $K_i+U_{gi}+U_{si}+W_{\text{tot}}=K_f+U_{gf}+U_{sf}+\Delta E_{\text{th}}$                                                              (I)

Here, $K_i$ is the initial kinetic energy, $U_{gi}$ is the initial gravitational potential energy, $U_{si}$ is the initial spring  potential energy, $W_{\text{tot}}$ is the total work done on the system, $K_f$ is the final kinetic energy, $U_{gf}$ is the final gravitational potential energy, $U_{sf}$ is the final spring potential energy, and $\Delta E_{\text{th}}$ is the change in thermal energy.

In this problem, Equation (I) will changes to (since all other energies are zero),

    $U_{gi}+U_{si}=\Delta E_{\text{th}}$                                                                                                        (II)

Write the expression for the initial gravitational potential energy.

    $U_{gi}=mgy$                                                                                                                (III)

Here, $m$ is the mass of the particle, $g$ is the gravitational constant, $y$ is the height at which the object is placed.

Write the expression for the initial potential energy of the spring.

    $U_{si}=\frac{1}{2}k{x}^2$                                                                                                              (IV)

Here, $k$ is the spring constant, $x$ is the compression distance.

Write the expression for the thermal energy.

    $\Delta E_{\text{th}}=F_{k}s$                                                                                                                (V)

Here, $F_k$ is the kinetic friction, $s$ is the total path length.

The total path length will be $s=x+d$, and use equation (V), (IV) and (III) in equation (II),

    $mgy+\frac{1}{2}k{x}^2=F_k\left(x+d\right)$                                                                                        (VI)

Kinetic friction is proportional to the normal force which equals the weight.

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

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

Write the expression for the normal force.

    $F_N=mg$                                                                                                               (VIII)

Use equation (VIII) in equation (VI),

    $F_k={\mu }_{k}mg$                                                                                                              (IX)

Use equation (VIII) in (VI), and solve for $d$.

    $\begin{array}{c}mgy+\frac{1}{2}k{x}^2={\mu }_{k}mg\left(x+d\right)\\ d=\frac{y}{{\mu }_k}+\frac{1}{2}\frac{k{x}^2}{{\mu }_{k}mg}-x\end{array}$                                                                            (VI)

Conclusion:

The displacement $y$ is found from the trigonometry of the arrangement.

    $\begin{array}{c}y=\left(1.8\text{m}\right)\sin 40°\\ =1.16\text{m}\end{array}$

Substitute $1.16\text{m}$ for $y$, $0.35$ for ${\mu }_k$, $345\text{N}/\text{m}$ for $k$, $0.30\text{m}$ for $x$, $9.81\text{m}/{\text{s}}^{}$ for $g$, $0.65\text{kg}$ for $m$ and $30.0\text{cm}$ for $x$ in equation (VI) to find $d$.

    $\begin{array}{c}d=\frac{1.16\text{m}}{0.35}+\frac{1}{2}\frac{\left(345\text{N}/\text{m}\right){\left(0.30\text{m}\right)}^2}{\left(0.35\right)\left(0.65\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)}-30.0\text{cm}\\ \text{=}\frac{1.16\text{m}}{0.35}+\frac{1}{2}\frac{\left(345\text{N}/\text{m}\right){\left(0.30\text{m}\right)}^2}{\left(0.35\right)\left(0.65\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)}-\left(30.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =10\text{m}\end{array}$

Therefore, the value of $d$ is $\underset{\_}{10\text{m}}$.