Chapter 8, Problem 77AP

(a)

To determine

The speed of the sled and rider at point C.

Answer to Problem 77AP

The speed of slader and rider at point C is $14.1\text{m/s}$.

Explanation of Solution

Consider sled-chute-earth as an isolated system. Since, there is no friction force acting on the system.

Write the eqution for conservation of energy from point A to point C.

Write the equation for conservation of energy

    $\Delta U+\Delta K=0$                                                                                               (I)

Here, $\Delta U$ is the change in potential energy and $\Delta K$ is the change in kinetic energy.

Since, the system has potential energy due to gravitation of earth.

Write the equation for gravitational potential energy.

    $U_g=mgh$

Here, $U$ is the potential energy due to gravity, $g$ is the acceleration due to gravity and $h$ is the height from surface of earth.

Write the expression for change in potential energy of the system.

    $\Delta U=U_f-U_i$

Here, $U_f$ is the final potential energy and $U_i$ is the initial potential energy.

Substitute $mg{h}_f$ for $U_f$ and $mg{h}_i$ for $U_i$ in above equation.

    $\Delta U=mg{h}_f-mg{h}_i$

Simplify the above equation.

    $\Delta U=mg\left(h_f-h_i\right)$                                                                                       (II)

Here, $h_f$ is the final height of the system from earth surface and $h_i$ is the initial height of the system from earth surface.

Since, the system has kinetic energy due to motion of sled and rider.

Write the equation for kinetic energy.

    $K=\frac{1}{2}m{v}^2$

Here, $K$ is the kinetic energy of the system, $m$ is the mass and $v$ is the speed of the system.

Write the expression for the change in kinetic energy.

    $\Delta K=K_f-K_i$

Here, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy.

Substitute $\frac{1}{2}m{v}_f{}^2$ for $K_f$ and $\frac{1}{2}m{v}_i{}^2$ for $K_i$ in above equation.

    $\Delta K=\left(\frac{1}{2}m{v}_f{}^2\right)-\left(\frac{1}{2}m{v}_i{}^2\right)$

Simplify the above equation.

    $\Delta K=\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)$                                                                                   (III)

Here, $v_f$ is the final speed of the system and $v_i$ is the initial speed of the system.

Substitute $\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)$ for $\Delta K$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (I).

    $mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)=0$

Rearrange the above equation for $v_f$.

    $v_f=\sqrt{v_i{}^2-2g\left(h_f-h_i\right)}$                                                                             (IV)

Conclusion:

Substitute $2.50\text{m/s}$ for $v_i$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $0\text{m}$ for $h_f$ and $9.76\text{m}$$h_i$ in equation (IV).

    $\begin{array}{c}{v}_f=\sqrt{{\left(2.50\text{m/s}\right)}^2-2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\left(0\text{m}\right)-\left(9.76\text{m}\right)\right)}\\ \text{=14}\text{.055}\text{m/s}\\ \approx \text{14}\text{.1m/s}\end{array}$

Thus, the speed of slader and rider at point C is $\underset{\_}{14.1\text{m/s}}$.

(b)

To determine

The magnitude of the total force the water exerts on the sled.

Answer to Problem 77AP

The magnitude of the total force the water exerts on the sled is $\underset{\_}{800\text{N}}$.

Explanation of Solution

Consider sled-water as a system.

Since, the friction force exerted by water is in retarding force and hence non-conservative force.

Write the equation for conservation of energy

    $\Delta U+\Delta K=W$                                                                                             (V)

Here, $W$ is the work done.

Write the equation for work done by retarding force.

    $W=-f_{k}d$

Here, $f_k$ is the friction for exerted by the water and $d$ is the distance travelled in opposite side of the force.

Substitute $-f_{k}d$ for $W$, $\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)$ for $\Delta K$ and $mg\left(h_f-h_i\right)$ for $\Delta U$ in equation (V).

    $mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)=-f_{k}d$

Rearrange the above equation for $f_k$.

    $f_k=-\frac{mg\left(h_f-h_i\right)+\frac{1}{2}m\left(v_f{}^2-v_i{}^2\right)}{d}$

Simplify the above equation.

    $f_k=-\frac{m}{2d}\left(2g\left(h_f-h_i\right)+\left(v_f{}^2-v_i{}^2\right)\right)$                                                          (VI)

Since, the total force exerted by the water is friction force and normal force acting on the sled.

Write the equation for normal force

    $n=mg$                                                                                                     (VII)

Here, $n$ is the normal force on the sled by the water, $m$ is the mass of system and $g$ is the acceleration due to gravity.

Write the expression for the magnitude of the total force

    $F=\sqrt{f_k{}^2+n^2+2n{f}_k\cos \theta }$

Here, $F$ is the magnitude of total force and $\theta$ is the angle between friction and normal force.

Since, the normal force and friction force are exerted in perpendicular direction.

Substitute $90°$ for $\theta$ in above equation.

    $F=\sqrt{f_k{}^2+n^2+2n{f}_k\cos 90°}$

Simplify the above equation.

    $F=\sqrt{f_k{}^2+n^2}$                                                                                         (VIII)

Conclusion:

Subsitute $80.0\text{kg}$ for $m$, $50.0\text{m}$ for $d$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $0\text{m}$ for $h_f$, $9.76\text{m}$ for $h_i$, $0\text{m/s}$ for $v_f$ and $2.50\text{m/s}$ for $v_i$ in equation (VI).

    $\begin{array}{c}{f}_k=-\frac{\left(80.0\text{kg}\right)}{\left(2\right)\left(50.0\text{m}\right)}\left(2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(0\text{m}-9.76\text{m}\right)+\left({\left(0\text{m/s}\right)}^2-{\left(2.50\text{m/s}\right)}^2\right)\right)\\ =-\left(0.8\text{kg}/\text{m}\right)\left(-197.546{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)\\ =158.04\text{N}\end{array}$

Subsitute $80.0\text{kg}$ for $m$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in equation (VII).

    $\begin{array}{l}n=\left(80.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =784\text{N}\end{array}$

Substitute $158.04\text{N}$ for $f_k$ and $784\text{N}$ for $n$ in equaiton (VIII)

    $\begin{array}{c}F=\sqrt{{\left(158.04\text{N}\right)}^2+{\left(784\text{N}\right)}^2}\\ \text{=799}\text{.77 N}\\ \approx \text{800 N}\end{array}$

Thus, the magnitude of the total force the water exerts on the sled is $800\text{N}$.

(c)

To determine

The magnitude of the force the chute exerts on the sled at point B.

Answer to Problem 77AP

The magnitude of the force the chute exerts on the sled at point B is $\underset{\_}{771\text{N}}$.

Explanation of Solution

Consider the sled on the chute at the point B as shown in figure (a).

Write the expression for the angle

    $\theta ={\sin }^{-1}\left(\frac{h}{l}\right)$                                                                                              (IX)

Here, $\theta$ is the angle as shown in the triangle in figure (I), $h$ is the height of the triangle and $l$ is the length of hypotenteous.

Since, there is no motion in perpendicular direction fo the motion of sled, hence the net force at the point B will be zero.

Write the expression for net force in y-direction as shown in figure (I).

    $\sum F_y=0$                                                                                                      (X)

Here, $\sum F_y$ is the net force in y -direction.

Conside the free body diagram of the sled and rider at point B.

Write the expression for net force.

    $\sum F_y=n-mg\cos \theta$

Substitute $n-mg\cos \theta$ for $\sum F_y$ in equation (X).

    $n-mg\cos \theta =0$

Rearrange the above equation for $n$.

    $n=mg\cos \theta$                                                                                            (XI)

Conclusion:

Substitute $9.76\text{m}$ for $h$ and $54.3\text{m}$ for $l$ in equation (IX).

    $\begin{array}{l}\theta ={\sin }^{-1}\left(\frac{9.76\text{m}}{54.3\text{m}}\right)\\ ={\sin }^{-1}\left(0.1797\right)\\ =10.3°\end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $10.3°$ for $\theta$ in equation (XI).

    $\begin{array}{c}n=\left(80.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\cos \left(10.3°\right)\\ =771.23\text{N}\\ \approx \text{771 N}\end{array}$

Thus, the magnitude of the force the chute exerts on the sled at point B is $\underset{\_}{771\text{N}}$.

(d)

To determine

The force exerted by the chute on the sled at point C where the chute is curving in the vertical plane.

Answer to Problem 77AP

The force exerted by the chute on the sled at point C where the chute is curving in the vertical plane is $\text{1}\text{.58}\times {\text{10}}^3\text{N}$.

Explanation of Solution

Consider the chute is curving in the vertical plane at point C.

The free body diagram of the sled at point C is as shown in figure (b).

Since, the sled has normal force and the centripital force is outward the center of curve.

Write the equation for net force at point C.

    $F_c+F_n+F_g=0$

Here, $F_c$ is the centripetal force, $F_n$ is the normal force and $F_g$ is the gravitational force.

Substitute  $-\frac{m{v}^2}{r}$ for $F_c$, $n$ for $F_n$ and $-mg$ for $F_g$ in above expression.

    $\left(-\frac{m{v}^2}{r}\right)+n+\left(-mg\right)=0$

Rearrange the above equation.

    $n=mg+\frac{m{v}^2}{r}$                                                                                           (XII)

Conclusion:

Subsitute $80.0\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $20.0\text{m}$ for $r$ and $14.1\text{m/s}$ for $v$ in equation (XII).

    $\begin{array}{l}n=\left(80.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)+\frac{\left(80.0\text{kg}\right){\left(14.1\text{m/s}\right)}^2}{20.0\text{m}}\\ =\left(784\text{N}+795.24\text{N}\right)\\ =1579.24\text{N}\\ \approx \text{1}\text{.58}\times {\text{10}}^3\text{N}\end{array}$

Thus, the force exerted by the chute on the sled at point C where the chute is curving in the vertical plane is $\text{1}\text{.58}\times {\text{10}}^3\text{N}$.