Chapter 8, Problem 13P

(a)

To determine

Whether the child-earth system is isolated or not.

Answer to Problem 13P

Therefore, the child-earth system is isolated as the only force that can do work on the child is her weight.

Explanation of Solution

The mass of the child is $m$, the height of the slide is $h$ and the height from where she launch into the air is $\frac{h}{5}$.

In the child-earth system, the total work on the child is done by the gravitation force (her weight) only as there is no friction and the air resistance is ignored. The work done by the normal force of the slide is zero because her displacement is perpendicular to the normal force.

Hence, the child-earth system is an isolated system.

Conclusion:

Therefore, the child-earth system is isolated as the only force that can do work on the child is her weight.

(b)

To determine

Whether there is a non-conservative force acting within the system or not.

Answer to Problem 13P

Therefore, there is no non conservative force acting within the system because there is no frictional force.

Explanation of Solution

Only the weight of the child does work whereas the normal force does no work since the displacement is always perpendicular to the normal force.

The child-earth system is isolated so, there is no loss of energy. There is no non- conservative force acting within the system because frictional force is absent here.

Conclusion:

Therefore, there is no non conservative force acting within the system because there is no frictional force.

(c)

To determine

The total energy of the earth-child system when the child is at the top of the water slide.

Answer to Problem 13P

The total energy of the earth-child system when the child is at the top of the water slide is $mgh$.

Explanation of Solution

Write the formula to calculate the total energy at the top of the water slide

    $E_{\text{t}}=P{E}_{\text{t}}+K{E}_{\text{t}}$        (I)

Here, $E_{\text{t}}$ is the total energy of the system at the top of the water slide, $P{E}_{\text{t}}$ is the potential energy of the system at the top of the water slide and $K{E}_{\text{t}}$ is kinetic energy of the system at the top of the water slide.

Write the formula to calculate the potential energy of the system at the top of the water slide

    $P{E}_{\text{t}}=mgh$

Here, $m$ is the mass of child, $g$ is the acceleration due to gravity and $h$ is the vertical maximum height of the slide.

As the child starts from rest therefore, the kinetic energy at the top is zero.

Substitute $0$ for $K{E}_{\text{t}}$ and $mgh$ for $P{E}_{\text{t}}$ in equation (I).

    $\begin{array}{c}{E}_{\text{t}}=0+mgh\\ =mgh\end{array}$

Conclusion:

Therefore, the total energy of the earth child system when the child is at the top of the water slide is $mgh$.

(d)

To determine

The expression for the total energy of the system at the launching point.

Answer to Problem 13P

The total energy of the system at the launching point is $\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$.

Explanation of Solution

The mass of the child is $m$, the height of the slide is $h$ and the height from where she launch into the air is $\frac{h}{5}$.

Write the formula to calculate the total energy at the launching point of earth child system

    $E_{\text{l}}=P{E}_{\text{l}}+K{E}_{\text{l}}$        (II)

Here, $E_{\text{l}}$ is the total energy of the system at the launching point of the water slide, $P{E}_{\text{l}}$ is the potential energy of the system at the launching point of the water slide and $K{E}_{\text{l}}$ is kinetic energy of the system at the launching point of the water slide.

Write the formula to calculate the potential energy of the system at the launching point of the water slide

    $P{E}_{\text{l}}=mg\frac{h}{5}$

Write the formula to calculate the kinetic energy of the system at the launching point of the water slide

    $K{E}_{\text{l}}=\frac{1}{2}m{v}_{\text{i}}^2$

Here, $v_{\text{i}}$ is the velocity of child at the launching point.

Substitute $\frac{1}{2}m{v}_{\text{i}}^2$ for $K{E}_{\text{l}}$ and $mg\frac{h}{5}$ for $P{E}_{\text{l}}$ in equation (II).

    $E_{\text{l}}=\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$

Conclusion:

Therefore, the total energy of the earth child system at the launching point is $\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$.

(e)

To determine

The expression for the total energy of the system at the highest point in her projectile motion.

Answer to Problem 13P

The total energy of the system at the highest point of the projectile motion is $\frac{1}{2}m{v}_{\text{xi}}^{\text{2}}+mg{y}_{\text{max}}$.

Explanation of Solution

Write the formula to calculate the total energy at the highest point in the projectile motion of earth child system

    $E_{\text{h}}=P{E}_{\text{h}}+K{E}_{\text{h}}$        (III)

Here, $E_{\text{h}}$ is the total energy of the system at the highest point of her projectile motion, $P{E}_{\text{h}}$ is the potential energy of the system at the highest point of her projectile motion and $K{E}_{\text{h}}$ is kinetic energy of the system at the highest point of her projectile motion.

Write the formula to calculate the potential energy of the system at the highest point of her projectile motion

    $P{E}_{\text{h}}=mg{y}_{\text{max}}$

Here, $y_{\max }$ is the vertical maximum height of her projectile motion.

Write the formula to calculate the kinetic energy of the system at the highest point of her projectile motion

    $\begin{array}{c}K{E}_{\text{h}}=\frac{1}{2}m{v}_{\text{h}}^2\\ =\frac{1}{2}m\left(v_{\text{xi}}^{\text{2}}+v_{\text{yi}}^{\text{2}}\right)\\ =\frac{1}{2}m\left(v_{\text{xi}}^{\text{2}}+0\right)\\ =\frac{1}{2}m{v}_{\text{xi}}^{\text{2}}\end{array}$

Here, $v_{\text{h}}$ is the velocity of child at the highest point of the projectile motion.

Substitute $\frac{1}{2}m{v}_{\text{xi}}^{\text{2}}$ for $K{E}_{\text{h}}$ and $mg{y}_{\text{max}}$ for $P{E}_{\text{h}}$ in equation (III).

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

Conclusion:

Therefore, the total energy of the system at the highest point of the projectile motion is $\frac{1}{2}m{v}_{\text{xi}}^{\text{2}}+mg{y}_{\text{max}}$.

(f)

To determine

The speed of child at launching point.

Answer to Problem 13P

The speed of child at launching point is $v_{\text{i}}=2\sqrt{\frac{2gh}{5}}$.

Explanation of Solution

As the total energy of the system remains conserved at every point therefore,

    $E_{\text{t}}=E_{\text{l}}$

Substitute $\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$ for $E_{\text{l}}$ and $mgh$ for $E_{\text{t}}$.

    $\begin{array}{c}mgh=\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}\\ \frac{1}{2}{v}_{\text{i}}^2=gh-\frac{gh}{5}\\ v_{\text{i}}=2\sqrt{\frac{2gh}{5}}\end{array}$

Conclusion:

Therefore, the speed of child at launching point is $v_{\text{i}}=2\sqrt{\frac{2gh}{5}}$.

(g)

To determine

The maximum airborne height $y_{\max }$ in terms of $h$ and the launch angle $\theta$.

Answer to Problem 13P

The maximum airborne height $y_{\max }$ in terms of $h$ and the launch angle $\theta$ is $\frac{h}{5}-\frac{4h{\sin }^2\theta }{5}$.

Explanation of Solution

Write the formula to calculate the maximum height $y_{\max }$ by conservation of energy, the total energy at the highest point is equal to the total energy at the launching point

    $E_{\text{h}}=E_{\text{l}}$

Substitute $\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$ for $E_{\text{l}}$ and $\frac{1}{2}m{v}_{\text{xi}}^2+mg{y}_{\text{max}}$ for $E_{\text{h}}$.

    $\frac{1}{2}m{v}_{\text{xi}}^2+mg{y}_{\text{max}}=\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}$        (IV)

The horizontal component of velocity $v_{\text{i}}$

    $v_{\text{xi}}=v_{\text{i}}\cos \theta$

Substitute $v_{\text{i}}\cos \theta$ for $v_{\text{xi}}$ in equation (IV).

    $\begin{array}{c}\frac{1}{2}m{\left(v_{\text{i}}\cos \theta \right)}^2+mg{y}_{\text{max}}=\frac{1}{2}m{v}_{\text{i}}^2+mg\frac{h}{5}\\ y_{\text{max}}=\frac{h}{5}-\frac{v_{\text{i}}{}^2{\sin }^2\theta }{2g}\end{array}$        (V)

Substitute $2\sqrt{\frac{2gh}{5}}$ for $v_{\text{i}}$ in equation (V).

    $\begin{array}{c}{y}_{\text{max}}=\frac{h}{5}+\frac{{\left(2\sqrt{\frac{2gh}{5}}\right)}^2{\sin }^2\theta }{2g}\\ =\frac{h}{5}-\frac{4h{\sin }^2\theta }{5}\end{array}$

Conclusion:

Therefore, the speed of child at launching point is $\frac{h}{5}-\frac{4h{\sin }^2\theta }{5}$.

(h)

To determine

Whether the answers would be same if the waterslide were not frictionless.

Answer to Problem 13P

Therefore, the answers would not be same if the waterslide were not frictionless.

Explanation of Solution

Due to the presence of frictional force, the total mechanical energy of the system would not conserve. There would be some frictional losses. So, the kinetic energy of the child at every point after the top of the slide would be less than the kinetic energy when there is no friction.

Thus, less kinetic energy means, her launch speed, maximum height and final speed would also be less.

Conclusion:

Therefore, the answers would not be same if the waterslide were not frictionless.