Chapter 8, Problem 8.77AP

Review. In 1887 in Bridgeport, Connecticut, C. J. Belknap built the water slide shown in Figure P8.77. A rider on a small sled, of total mass 80.0 kg, pushed off to start at the top of the slide (point Ⓐ) with a speed of 2.50 m/s. The chute was 9.76 m high at the top and 543 m long. Along its length, 72.5 small wheels made friction negligible. Upon leaving the chute horizon-tally at its bottom end (point ©), the rider skimmed across the water of Long Island Sound for as much as 50 m, “skipping along like a flat pebble,” before at last coming to rest and swimming ashore, pulling his sled after him. (a) Find the speed of the sled and rider at point © (b) Model the force of water friction as a constant retarding force acting on a particle. Find the magnitude of the friction force the water exerts on the sled. (c) Find the magnitude of the force the chute exerts on the sled at point Ⓑ (d) At point ©, the chute is horizontal but curving in the vertical plane. Assume its radius of curvature is 20.0 m. Find the force the chute exerts on the sled at point ©.

To determine

The speed of the sled and rider at point $C$.

Answer to Problem 8.77AP

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

Explanation of Solution

Given info: The speed at point $A$ is $2.50\text{m}/\text{s}$, height of chute is $9.76\text{m}$, length of chute is $54.3\text{m}$, radius of curvature of chute is $20.0\text{m}$, mass of the rider is $80.0\text{kg}$.

The formula to calculate the kinetic energy at point $A$ is,

$K_{\text{A}}=\frac{1}{2}m{v}_{\text{A}}{}^2$

Here,

$m$ s the mass of the rider.

$v_{{}_{\text{A}}}$ is the speed at point $A$.

The formula to calculate the initial gravitational potential energy at point $A$ is,

$U_{\text{A}}=mg{h}_{\text{A}}$

Here,

$m$ is the mass of the rider.

$g$ is the acceleration due to gravity.

$h_{\text{A}}$ is the height at point $A$.

The formula to calculate the gravitational potential energy at point $C$ is,

$U_{\text{C}}=mg{h}_{\text{C}}$

Here,

$m$ is the mass of the rider.

$g$ is the acceleration due to gravity.

$h_{\text{C}}$ is the height at point $C$

The formula to calculate the kinetic energy at point $C$ is,

$K_{\text{C}}=\frac{1}{2}m{v}_{\text{C}}{}^2$

Here,

$m$ is the mass of the rider.

$v_{\text{C}}$ is the velocity at point $C$.

The formula to calculate the energy at point $A$ is,

$E_{\text{A}}=K_{\text{A}}+U_{\text{A}}$

Here,

$K_{\text{A}}$ is the kinetic energy at point $A$

$U_{\text{A}}$ is the gravitational potential energy at point $A$.

Substitute $mg{h}_{\text{A}}$ for $U_{\text{A}}$ and $\frac{1}{2}m{v}_{\text{A}}{}^2$ for $K_{\text{A}}$ to find $E_{\text{A}}$.

$E_{\text{A}}=mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2$

The formula to calculate the energy at point $C$ is,

$E_{\text{C}}=K_{\text{C}}+U_{\text{C}}$

Here,

$K_{\text{C}}$ is the final kinetic energy of the block.

$U_{\text{C}}$ is the final potential energy of the block

Substitute $mg{h}_{\text{c}}$ for $U_{\text{C}}$ and $\frac{1}{2}m{v}_{\text{C}}{}^2$ for $K_{\text{f}}$ to find $E_{\text{C}}$.

$E_{\text{C}}=\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}$

Apply the law of conservation of energy at point $A$ and $C$.

$E_{\text{A}}=E_{\text{C}}$

Here,

$E_{\text{A}}$ is the energy at point $A$

$E_{\text{C}}$ is the energy at point $C$.

Substitute $\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}$ for $E_{\text{C}}$ and $mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2$ for $E_{\text{A}}$ in the above formula.

$mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2=\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}$

Rearrange the above formula for $v_{\text{C}}$.

$\begin{array}{c}{v}^2{}_{\text{A}}+2g{h}_{\text{A}}=v^2{}_{\text{C}}+2g{h}_{\text{C}}\\ v^2{}_{\text{C}}=v^2{}_{\text{A}}+2g{h}_{\text{A}}-2g{h}_{\text{C}}\end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.50\text{m}/\text{s}$ for $v_{\text{A}}$, $0$ for $h_{\text{C}}$, $250\text{N}/\text{m}$ for $k$, $9.76\text{m}$ for $h_{\text{A}}$ in the above formula to find $v_{\text{C}}$.

$\begin{array}{c}{v}^2{}_{\text{C}}={\left(2.50\text{m}/\text{s}\right)}^2+2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(9.76\text{m}\right)-2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0\right)\\ v_{\text{C}}=\sqrt{{\left(2.50\text{m}/\text{s}\right)}^2+2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(9.76\text{m}\right)}\\ =\sqrt{192.546{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =14.055\text{m}/\text{s}\end{array}$

Take the approximation.

$v_{\text{C}}\approx 14.1\text{m}/\text{s}$

Conclusion:

Therefore, the speed of the sled and rider at point $C$ is $14.1\text{m}/\text{s}$.

To determine

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

Answer to Problem 8.77AP

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

Explanation of Solution

Given info: The speed at point $A$ is $2.50\text{m}/\text{s}$, height of chute is $9.76\text{m}$, length of chute is $54.3\text{m}$, radius of curvature of chute is $20.0\text{m}$, mass of the rider is $80.0\text{kg}$.

The free body diagram is shown below.

Figure II

The formula to calculate the work done by the friction force at point $A$ is,

$W=f_{\text{A}}s$

Here,

$f_{\text{A}}$ is the friction force at point $A$.

$s$ is the displacement.

The formula to calculate the energy at point $A$ is,

$E_{\text{A}}=K_{\text{A}}+U_{\text{A}}-W$

Here,

$K_{\text{A}}$ is the kinetic energy at point $A$

$U_{\text{A}}$ is the gravitational potential energy at point $A$.

$W$ is the work done by the frictional force.

Substitute $mg{h}_{\text{A}}$ for $U_{\text{A}}$, $\frac{1}{2}m{v}_{\text{A}}{}^2$ for $K_{\text{A}}$, $f_{\text{A}}s$ for $W$ to find $E_{\text{A}}$.

$E_{\text{A}}=mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2-f_{\text{A}}s$

The formula to calculate the energy at point $C$ is,

$E_{\text{C}}=K_{\text{C}}+U_{\text{C}}$

Here,

$K_{\text{C}}$ is the final kinetic energy of the block.

$U_{\text{C}}$ is the final potential energy of the block

Substitute $mg{h}_{\text{C}}$ for $U_{\text{C}}$ and $\frac{1}{2}m{v}_{\text{C}}{}^2$ for $K_{\text{f}}$ to find $E_{\text{C}}$.

$E_{\text{C}}=\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}$

Apply the law of conservation of energy at point $A$ and $C$.

$E_{\text{A}}=E_{\text{C}}$

Here,

$E_{\text{A}}$ is the energy at point $A$

$E_{\text{C}}$ is the energy at point $C$.

Substitute $\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}$ for $E_{\text{C}}$ and $mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2-f_{\text{A}}s$ for $E_{\text{A}}$ in the above formula.

$mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2=\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}-f_{\text{A}}s$

Rearrange the above formula for $f_{\text{A}}s$.

$\begin{array}{c}mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2=\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}-f_{\text{s}}s\\ f_{\text{A}}s=mg{h}_{\text{A}}+\frac{1}{2}m{v}_{\text{A}}{}^2-\left(\frac{1}{2}m{v}_{\text{C}}{}^2+mg{h}_{\text{C}}\right)\end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.50\text{m}/\text{s}$ for $v_{\text{A}}$, $0$ for $h_{\text{C}}$, $250\text{N}/\text{m}$ for $k$, $9.76\text{m}$ for $h_{\text{A}}$,0 for $v_{\text{C}}$  in the above formula to find $f_{\text{A}}s$.

$\begin{array}{c}{f}_{\text{A}}s=\left(80.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(9.76\text{m}\right)+\frac{1}{2}\left(80.0\text{kg}\right){\left(2.50\text{m}/\text{s}\right)}^2-\left(\begin{array}{l}\frac{1}{2}\left(80.0\text{kg}\right){\left(0\right)}^2+\\ \left(80.0\text{kg}\right)\\ \left(\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0\right)\right)\end{array}\right)\\ =7901.84\text{J}\\ \text{=7}\text{.90}\times {10}^3\text{J}\end{array}$

Thus, the value of work done by the frictional force is $\text{7}\text{.90}\times {10}^3\text{J}$.

From the above figure, the displacement is $50.0\text{m}$.

Substitute $\text{7}\text{.90}\times {10}^3\text{J}$ for $W$ and $50.0\text{m}$ for $s$ in the formula (1)  to find $f_{\text{s}}$.

$\begin{array}{c}\text{7}\text{.90}\times {10}^3\text{J}=f_{\text{A}}\left(50.0\text{m}\right)\\ f_{\text{A}}=\frac{\text{7}\text{.90}\times {10}^3\text{J}}{50.0\text{m}}\\ =0.158\times {10}^3\text{N}\\ \text{=158}\text{N}\end{array}$

Thus, the frictional force acts on point $A$ is $158\text{N}$.

The formula to calculate the normal force is,

$N=mg$

Here,

$m$ is the mass of the rider.

$g$ is the acceleration due to gravity.

Substitute $80.0\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above formula to find $N$.

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

Thus, the value of normal force is $784\text{N}$.

The formula to calculate the magnitude of the force the water exerts on the sled is,

$F=\sqrt{{\left(N\right)}^2+{\left(f_{\text{A}}\right)}^2}$

Here,

$N$ is the normal force.

$f_{\text{A}}$ is the frictional force acting at point $A$.

Substitute $158\text{N}$ for $f_{\text{A}}$ and $784\text{N}$ for $N$ in the above formula to find $F$.

$\begin{array}{c}F=\sqrt{{\left(784\text{N}\right)}^2+{\left(158\text{N}\right)}^2}\\ =\sqrt{639320\text{N}}\\ =799.76\text{}\text{N}\\ \approx \text{800}\text{N}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 8.77AP

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

Explanation of Solution

Given info: The speed at point $A$ is $2.50\text{m}/\text{s}$, height of chute is $9.76\text{m}$, length of chute is $54.3\text{m}$, radius of curvature of chute is $20.0\text{m}$, mass of the rider is $80.0\text{kg}$.

From the above figure,

$\begin{array}{c}\sin \theta =\frac{9.76\text{m}}{54.3\text{m}}\\ =0.1794\\ \theta ={\sin }^{-1}\left(0.1794\right)\\ =10.35°\end{array}$

The formula to calculate the force exerted by the weight of the chute is,

$W=mg\cos \theta$

Here,

$m$ is the mass of the chute.

$g$ is the acceleration due to gravity.

The formula to calculate the force acting at point $B$ is,

$F=ma$

Here,

$m$ is the mass of the chute.

$a$ is the acceleration.

The acceleration is 0 at point $B$ then the net force acting at point $B$ is 0

The formula to calculate the force at the point $B$ is,

$N-W=F$

Here,

$N$ is the normal force acting at point $B$

$W$ is the weight of the sled.

$F$ is the net force acting at point $B$.

Substitute $mg\cos \theta$ for $W$,0 for $F$ in the above formula to find $N$.

$\begin{array}{c}N-W=F\\ N-mg\cos \theta =0\\ N=mg\cos \theta \end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10.35°$ for $\theta$ in the above formula to find $N$.

$\begin{array}{c}N=\left(80.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\cos 10.35°\\ =\left(80.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0.983\right)\\ =770.672\text{}\text{N}\\ \text{=771}\text{N}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 8.77AP

The magnitude of the force the chute exerts on the sled at point $B$ is $\text{1}\text{.57}\text{}\text{kN}$.

Explanation of Solution

Given info: The speed at point $A$ is $2.50\text{m}/\text{s}$, height of chute is $9.76\text{m}$, length of chute is $54.3\text{m}$, radius of curvature of chute is $20.0\text{m}$, mass of the rider is $80.0\text{kg}$.

From the above figure,

$\begin{array}{c}\sin \theta =\frac{9.76\text{m}}{54.3\text{m}}\\ =0.1794\\ \theta ={\sin }^{-1}\left(0.1794\right)\\ =10.35°\end{array}$

The formula to calculate the centripetal force exerted at point $C$ is,

$F_{\text{C}}=\frac{m{v}_{\text{C}}}{r}$

Here,

$m$ is the mass of the chute.

$v_{\text{C}}$ is the speed at point $C$

The formula to calculate the weight of chute at point $C$ is,

$W=mg$

Here,

$m$ is the mass of the chute.

$g$ is the acceleration due to gravity.

The formula to calculate the force at the point $C$ is,

$N_{\text{C}}-W=F_{\text{C}}$

Here,

$N$ is the normal force acting at point $C$

$W$ is the weight of the sled.

$F$ is the centripetal force acting at point $C$.

Substitute $mg$ for $W$, $\frac{m{v}_{\text{C}}}{r}$ for $F_{\text{C}}$ in the above formula to find $N_{\text{C}}$.

$\begin{array}{c}{N}_{\text{C}}-mg=\frac{m{v}_{\text{C}}}{r}\\ N=mg+\frac{m{v}_{\text{C}}}{r}\end{array}$

Substitute $80.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10.41\text{m}/\text{s}$ for $v_{\text{C}}$ in the above formula to find $N_{\text{C}}$.

$\begin{array}{c}{N}_{\text{C}}=\left(80.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)+\frac{\left(80.0\text{kg}\right)\left(14.1\text{m}/\text{s}\right)}{\left(20.0\text{m}\right)}\\ =784\text{N+795}\text{.24}\text{}\text{N}\\ \text{=1579}\text{.24}\text{N}\\ \approx \text{1}\text{.57}\text{}\text{kN}\end{array}$

Conclusion:

Therefore, the magnitude of the force the chute exerts on the sled at point $B$ is $\text{1}\text{.57}\text{}\text{kN}$.