Chapter 9, Problem 77PQ

(a)

To determine

The speed of the block when it reaches the bottom of the curve.

Answer to Problem 77PQ

The speed of the block when it reaches the bottom of the curve is $\underset{\_}{3.3\text{m}/\text{s}}$

Explanation of Solution

The energy conservation equation for a system is,

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

Here, $K_i$ is the initial kinetic energy, $U_{gi}$ is the initial gravitational potential energy, $U_{ei}$ is the initial elastic 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, $U_{ef}$ is the final elastic potential energy, and $\Delta E_{\text{th}}$ is the change in thermal energy.

There is no friction during the motion so that the change in thermal energy will also be zero ($\Delta E_{\text{th}}=0$),also there are no identifiable forces performing work on the system, other than gravity so the total work done on the system is zero ($W_{\text{tot}}=0$). Since only conservative forces acting on the system, equation (I) can be reduced to,

  $K_i+U_{gi}+U_{ei}=K_f+U_{gf}+U_{ef}$                                                                                (II)

The total potential energy is the sum of the gravitational potential energy and the elastic potential energy.

  $\begin{array}{l}{U}_i=U_{gi}+U_{ei}\\ U_f=U_{gf}+U_{ef}\end{array}$

Using the expression for the total potential energy of the system in the initial and final condition to equation (II) yields,

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

The initial kinetic energy of the block is zero and at the bottom of the track where $y=0$, yields zero final potential energy to the block. Thus equation (IV) can be reduced to,

  $\begin{array}{c}0+U_i=K_f+0\\ U_i=K_f\end{array}$                                                                                                        (V)

Write the expression for the final kinetic energy of the block.

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

Here, $m$ is the mass of the block, $v$ is the speed of the block.

Write the expression for the initial gravitational potential energy of the block.

    $U_i=mgh$                                                                                                               (VII)

Here, $g$ is the acceleration due to gravity, $h$ is the height of the object.

use equation (VI) and (VII) in equation (V) and solve for $v_f$,

    $\begin{array}{c}mgh=\frac{1}{2}m{v}_f^2\\ v_f^2=2gh\\ v_f=\sqrt{2gh}\end{array}$                                                                                                        (VIII)

Conclusion:

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$, $0.54\text{m}$ for $h$ in equation (VIII) to find $v_f$.

    $\begin{array}{c}{v}_f=\sqrt{2\left(9.81\text{m}/{\text{s}}^2\right)\left(0.54\text{m}\right)}\\ =3.3\text{m}/\text{s}\end{array}$

Therefore, the speed of the block when it reaches the bottom of the curve is $\underset{\_}{3.3\text{m}/\text{s}}$

(b)

To determine

The magnitude of the friction force acting on the block.

Answer to Problem 77PQ

The magnitude of the friction force acting on the block is $1.8\text{N}$.

Explanation of Solution

Applying the conservation of energy condition from the moment the block reaches the bottom of the track until it finally stops. While there are no external forces performing work on the block-track system during this motion ($W_{\text{tot}}=0$), friction will cause a rise in internal energy. The final velocity will be zero, so that equation (I) changes to,

    $K_i=\Delta E_{\text{th}}$                                                                                                                (IX)

The initial kinetic energy in the horizontal path in the bottom is equal to the final kinetic energy calculated in part (a) that corresponds to a speed of $3.3\text{m/s}$.

Modify equation (IX) using the expression for the kinetic energy.

    $\frac{1}{2}m{v}_i^2=\Delta E_{\text{th}}$                                                                                                            (X)

Write the expression for the change in thermal energy.

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

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

Use equation (X) in equation (XI), and solve for $F_k$.

    $\begin{array}{c}\frac{1}{2}m{v}_i^2=F_{k}s\\ F_k=\frac{m{v}_i^2}{2s}\end{array}$                                                                                                         (XII)

Conclusion:

Substitute $0.25\text{kg}$ for $m$, $3.3\text{m}/\text{s}$ for $v$, $0.75\text{m}$ for $s$ in equation (XII) to find $F_k$.

    $\begin{array}{c}{F}_k=\frac{\left(0.25\text{kg}\right){\left(3.3\text{m}/\text{s}\right)}^2}{\left(2\right)\left(0.75\text{m}\right)}\\ =1.8\text{N}\end{array}$

Therefore, the magnitude of the friction force acting on the block is $1.8\text{N}$.