Chapter 7, Problem 77P

(a)

To determine

Speed of the sled and rider.

Answer to Problem 77P

Speed is $\underset{\_}{14.\text{1}\text{m}/\text{s}}$.

Explanation of Solution

Write down the equation of energy conservation.

  $K_i+U_{gi}=K_f+U_{gf}$        (I)

Here $K_i$ is the initial kinetic energy, $U_{gi}$ is the initial gravitational potential energy, $K_f$ is the final kinetic energy and $U_{gf}$ is the final gravitational potential energy

Write the equation for initial and final kinetic energy

  $\begin{array}{c}{K}_i=\frac{1}{2}m{v}_A{}^2\\ K_f=\frac{1}{2}m{v}_C{}^2\end{array}$        (II)

Here $m$ is the mass, $v_A$ is the velocity at point A and $v_C$ is the velocity at point C.

Write down the equation for initial and final gravitational potential energy.

  $\begin{array}{l}{U}_{gi}=mg{y}_A\\ U_{gf}=0\end{array}$        (III)

Here $g$ is the acceleration due to gravity and $y_A$ is the height at A.

Substitute (II) and (III) in (I)

  $\frac{1}{2}m{v}_A{}^2+mg{y}_A=\frac{1}{2}m{v}_C{}^2+0$        (IV)

Rearrange (IV) in terms of $v_C$.

    $\begin{array}{c}\frac{1}{2}m{v}_C{}^2=\frac{1}{2}m{v}_A{}^2+mg{y}_A\\ v_C{}^2=v_A{}^2+2g{y}_A\\ v_C=\sqrt{v_A{}^2+2g{y}_A}\end{array}$        (V)

Conclusion:

Substitute $2.5\text{m}/\text{s}$ for $v_A$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $9.76\text{m}$ for $y_A$ in (V)

    $\begin{array}{c}{v}_C=\sqrt{{\left(2.5\text{m}/\text{s}\right)}^2+2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(9.76\text{m}\right)}\\ =14.1\text{m}/\text{s}\end{array}$

Speed is $\underset{\_}{14.\text{1}\text{m}/\text{s}}$.

(b)

To determine

Magnitude of force on water

Answer to Problem 77P

Force is $\underset{\_}{800\text{N}}$.

Explanation of Solution

Write the energy conservation equation for a motion incorporated with frictional force

    $K_i+U_{gi}-f_{k}d=K_f+U_{gf}$        (VI)

Here $f_k$ is the frictional force and $d$ is the distance covered.

Substitute (II) and (III) in (VI)

    $\frac{1}{2}m{v}_A{}^2+mg{y}_A-f_{k}d=\frac{1}{2}m{v}_C{}^2+0$        (VII)

Final kinetic energy is also zero as the final velocity is zero.

Now rearrange (VII) in terms of $f_k$

    $f_k=\frac{\frac{1}{2}m{v}_A{}^2+mg{y}_A}{d}$        (VIII)

Write the equation for normal force acting

    $n=mg$        (IX)

Therefore total force on water is,

    $F=\sqrt{f_k{}^2+n^2}$        (X)

Here $F$ is the magnitude of total force on water.

Conclusion:

Substitute $80\text{kg}$ for $m$, $2.5\text{m}/\text{s}$ for $v_A$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $50\text{m}$ for $d$ and $9.76\text{m}$ for $y_A$ in (VIII)

    $\begin{array}{c}{f}_k=\frac{\frac{1}{2}\left(80\text{kg}\right){\left(2.5\text{m}/\text{s}\right)}^2+\left(80\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(9.76\text{m}\right)}{50\text{m}}\\ =158\text{N}\end{array}$

Substitute $80\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (IX)

    $\begin{array}{c}n=\left(80\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =784\text{N}\end{array}$

Substitute $158\text{N}$ for $f_k$ and $784\text{N}$ for $n$ in (X)

    $\begin{array}{c}F=\sqrt{{\left(158\text{N}\right)}^2+{\left(784\text{N}\right)}^2}\\ =800\text{N}\end{array}$

Force is $\underset{\_}{800\text{N}}$.

(c)

To determine

Force exerted by the chute.

Answer to Problem 77P

Force is $\underset{\_}{771\text{N}}$.

Explanation of Solution

Use Pythagoras theorem to get the angle of slide

    $\begin{array}{c}\sin \theta =\frac{\text{opp}}{\text{hyp}}\\ \theta ={\sin }^{-1}\left(\frac{\text{opp}}{\text{hyp}}\right)\end{array}$        (XI)

Here $\theta$ is the angle of slide, $\text{opp}$ is the side opposite to theta, and $\text{hyp}$ is the hypotenuse of the right triangle.

Write the equation for force on $B$

  $n=mg\cos \theta$        (XII)

Conclusion:

Substitute $9.76\text{m}$ for $\text{opp}$, $53.4\text{m}$ for $\text{hyp}$ in (XI)

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{9.76\text{m}}{54.3\text{m}}\right)\\ =10.4°\end{array}$

Substitute $80\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10.4°$ for $\theta$ in (XIII)

    $\begin{array}{c}n=\left(80\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\cos 10.4°\\ =771\text{N}\end{array}$

Force is $\underset{\_}{771\text{N}}$.

(d)

To determine

Force exerted by chute on point C.

Answer to Problem 77P

The force is $\underset{\_}{1.57\times {10}^3\text{N}}$.

Explanation of Solution

Write the equation for net force.

    $F_C=n_C-mg$        (XIII)

Here $n_C$ is the force at C.

Write the equation for centripetal force

    $F_C=\frac{m{v}_C{}^2}{r}$        (XIV)

Here $r$ is the radius of the circular path.

Equate (XIII) and (XIV) and write for $n_C$.

    $\begin{array}{c}{n}_C-mg=\frac{m{v}_C{}^2}{r}\\ n_C=mg+\frac{m{v}_C{}^2}{r}\end{array}$        (XV)

Conclusion:

Substitute $80\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $14.1\text{m}/\text{s}$ for $v_C$ and $20\text{m}$ for $r$ in (XV)

    $\begin{array}{c}{n}_C=\left(80\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)+\frac{\left(80\text{kg}\right){\left(14.1\text{m}/\text{s}\right)}^2}{20\text{m}}\\ =1.57\times {10}^3\text{N}\end{array}$

The force is $\underset{\_}{1.57\times {10}^3\text{N}}$.