Question

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 A) with a speed of 2.50 m/s. The chute was 9.76 m high at the top and 54.3 m long. Along its length, 725 small wheels made friction negligible. Upon leaving the chute horizontally at its bottom end (point C), 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 C.

(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 B.

(d) At point C, 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 C.

Short answer

(a) The speed of rider and sled at point C is $14.1\text{\hspace{0.17em}m}/\text{s}$.

(b) The magnitude of the friction force exerted on the sled by the water is $159.2\text{\hspace{0.17em}N}$.

(c) The force exerted by the chute on the sled at point B is $771\text{\hspace{0.17em}N}$.

(d) The force exerted by the chute on the sled at point C is $1.57\text{\hspace{0.17em}kN}$.

Step-by-step solution

Identification of given data

The height of chute from the bottom is $\mathrm{h}=9.76\text{\hspace{0.17em}m}$.

The length of chute is $l=54.3\text{\hspace{0.17em}m}$.

The speed of the sled at point A is $\mathrm{u}=2.50\text{\hspace{0.17em}\hspace{0.17em}m/s}$

The distance covered by the rider on water is $\mathrm{d}=50\text{\hspace{0.17em}m}$.

The total mass of the sled is $\mathrm{m}=80\text{\hspace{0.17em}kg.}$

The radius of curve at point C is $\mathrm{R}=20\text{\hspace{0.17em}m.}$

Concept

The principle of energy conservation for an isolated system states that the sum of the change in potential energy, change in kinetic energy, and change in internal energy of a system should be zero. This means that the total energy of a system remains constant.

$\mathbf{\Delta U}\mathbf{+}\mathbf{\Delta K}\mathbf{+}{\mathbf{\Delta E}}_i\mathbf{=}\mathbf{0}$

Here,$\mathbf{U}$ is the potential energy,$\mathbf{K}$ is the kinetic energy, and${\mathbf{E}}_i$ is the internal energy.

Determination of speed of rider and sled at point C(a)

The speed of rider and the sled at point C is given as:

${\mathrm{v}}_{\mathrm{C}}^2={\mathrm{u}}^2+2\mathrm{gh}$

Replace all the values in the above equation.

$\begin{array}{l}{\mathrm{v}}_{\mathrm{C}}^2={(2.50\text{\hspace{0.17em}m}/\text{s})}^2+2(10\text{\hspace{0.17em}m}/{\text{s}}^2)\left(9.76\text{\hspace{0.17em}m}\right)\\ {\mathrm{v}}_{\mathrm{C}}=14.1\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, the speed of the rider and the sled at point C is $14.1\text{\hspace{0.17em}m}/\text{s}$.

Determination of magnitude of friction exerted by water on sled(b)

Use the third equation of motion for the motion of the sled on the water. For the initial velocity $\left(\mathrm{u}\right)=14.1\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}$, the final velocity role="math" localid="1663738143107" $\left(v\right)=0$, and the distance travelled $\left(\mathrm{s}\right)=50\text{\hspace{0.17em}m}$, the equation can be written as:

$\begin{array}{c}{\mathrm{v}}^2={\mathrm{u}}^2-2\mathrm{as}\\ 0={(14.1\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2-2\mathrm{a}\left(50\text{\hspace{0.17em}m}\right)\\ \mathrm{a}=\frac{{(14.1\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2}{2\left(50\text{\hspace{0.17em}m}\right)}\\ =1.99{\text{m}/\text{s}}^2.\end{array}$

As friction is the only responsible factor for the retardation of the sled, the frictional force can be calculated as:

$\begin{array}{c}\mathrm{f}=\mathrm{ma}\\ =\left(80\text{\hspace{0.17em}kg}\right)(1.99\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})\\ =159.2\text{\hspace{0.17em}N.}\end{array}$

Therefore, the magnitude of the friction force exerted on the sled by the water is $159.2\text{\hspace{0.17em}N}$.

Determination of force exerted by the chute on sled at point B(c)

The slope of the chute is given as:

$\begin{array}{c}\mathrm{sin\theta }=\frac{\mathrm{h}}{\mathrm{l}}\\ \mathrm{sin\theta }=\frac{9.76\text{\hspace{0.17em}m}}{54.3\text{\hspace{0.17em}m}}\\ \mathrm{\theta }=10.4°\end{array}$

The force exerted by the chute on the sled at point B is given as:

${\mathrm{F}}_{\mathrm{c}}=\mathrm{mgcos\theta }$

Replace all the values in above equation.

$\begin{array}{c}{\mathrm{F}}_{\mathrm{c}}=\left(80\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)(\cos 10.4°)\\ =771\text{\hspace{0.17em}N.}\end{array}$

Therefore, the force exerted by the chute on the sled is $771\text{\hspace{0.17em}N}$.

Determination of force exerted by the chute on sled at point C(d)

The force exerted by the chute on the sled at point C is given as:

$\mathrm{F}=\mathrm{m}\left(\frac{{\mathrm{v}}_{\mathrm{C}}^2}{\mathrm{R}}+\mathrm{g}\right)$

Replace all the values in above equation.

$\begin{array}{c}\mathrm{F}=\left(80\text{\hspace{0.17em}kg}\right)\left[\frac{{(14.1\text{\hspace{0.17em}m}/\text{s})}^2}{20\text{\hspace{0.17em}m}}+9.8\text{\hspace{0.17em}m}/{\text{s}}^2\right]\\ =1579.24\text{\hspace{0.17em}N}\\ =\left(1579.24\text{\hspace{0.17em}N}\right)\left(\frac{1\text{\hspace{0.17em}kN}}{1000\text{\hspace{0.17em}N}}\right)\\ =1.57\text{\hspace{0.17em}kN.}\end{array}$

Therefore, the force exerted by the chute on the sled at point C is $1.57\text{\hspace{0.17em}kN}$.