Question

A child lying on her back experiences $\mathbf{55}\mathbf{.}\mathbf{0}\mathit{N}$tension in the muscles on both sides of her neck when she raises her head to look past her toes. Later, sliding feet first down a water slide at terminal speed $\mathbf{5}\mathbf{.}\mathbf{70}\mathit{m}\mathbf{/}\mathit{s}$s and riding high on the outside wall of a horizontal curve of radius,$\mathbf{2}\mathbf{.}\mathbf{40}\mathit{m}$she raises her head again to look forward past her toes. Find the tension in the muscles on both sides of her neck while she is sliding.

Short answer

The tension in the muscles is$F\text{'}=93.8N$

Step-by-step solution

of 3: Definition of the acceleration 

When a body is moving in a circular motion, its horizontal acceleration is equal to,

$a_x=\frac{v^2}{r}$

The terminal speed is,$v$while the horizontal curve radius is $r$.

The vertical acceleration is the same as the gravitational acceleration.$a_y=g$

Thus, the net acceleration is

$\begin{array}{c}a=\sqrt{{a_x}^2+{a_g}^2}\\ =\sqrt{{\left(\frac{v^2}{r}\right)}^2+g^2}\mathrm{................................}\left(1\right)\end{array}$

of 3: Determine the net acceleration and the factor f

The child's acceleration when sliding is.

$a=\sqrt{{\left(\frac{v^2}{r}\right)}^2+g^2}$

Plugging the child's terminal speed,$v=5.70m/s,r=2.40m$ Horizontal radius curve We get $g=9.8m/s^2$due to gravitational acceleration,

$\begin{array}{c}a=\sqrt{a_x^2+a_g^2}\\ =\sqrt{{\left(\frac{v^2}{r}\right)}^2+g}\\ =\sqrt{{\left(\frac{{(5.70m/s)}^2}{(2.40m)}\right)}^2+{(9.8m/s^2)}^2}\\ =16.7m/s^2\end{array}$

This is large than $g$by a factor,

$\begin{array}{c}f=\frac{a}{g}\\ =\frac{(16.7m/s^2)}{(9.8m/s^2)}\\ =1.71\end{array}$

of 3: Determine the tenson 

$F=55.0N$, muscle tension on both sides of her neck.

As a result, she has tension in the muscles on both sides of her neck as she slides.

$\begin{array}{c}F\text{'}=Ff\\ =(55N)(1.71)\\ =93.8N\end{array}$