A child of mass 26kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 12m/sAt this instant the cord is 4.30mlong.

(a) At this instant, what is the parallel component of the rate of change of the child's momentum?

(b) At this instant, what is the perpendicular component of the rate of change of the child's momentum?

(c) At this instant, what is the net force acting on the child?

(d) What is the magnitude of the force that the elastic cord exerts on the child? (It helps to draw a diagram of the forces.)

(e) The relaxed length of the elastic cord is 4.22m. What is the stiffness of the cord?

A child of mass $m=26\mathrm{kg}$, swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal and the speed is$v=12\mathrm{m}/\mathrm{s}$. At this instant the cord is $R=4.30\mathrm{m}$long.

A force is a push or pull on an object is because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them .The acted force may be of attraction or of repulsion .The two items no longer feel the force after the interaction ends

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum$\frac{d\overrightarrow{p}}{dt}$and the perpendicular rate of change of momentum$\frac{d\overrightarrow{p}}{dt}$are the two elements that we are concerned with.

So, the net force$F_{net}$on the object is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =\frac{d\overrightarrow{p}}{dt}+\frac{d\overrightarrow{p}}{dt} \end{gathered}$$

The rock's speed is affected by the parallel rate of change of momentum and given the speed is constant, therefore, the parallel rate is zero, and it equals the size of the momentum rate change.

$\frac{d\overrightarrow{p}}{dt}=0$

Thus, the parallel component of the rate of change of the child's momentum is zero.

The rate change is the direction shift caused by the perpendicular rate of change.

At speeds far slower than the speed of light, the amount of the perpendicular rate change matches the rate change of the direction of the momentum.

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

Now put the values for$m$,$v$and$R$to getrole="math" localid="1656908974157" $\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt}$

role="math" localid="1656909087866" $$\begin{gathered} \frac{m{v}^2}{R}=\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt} \\ =\frac{\left(26\mathrm{kg}\right){\left(12\mathrm{m}/\mathrm{s}\right)}^2}{4.3\mathrm{m}} \\ =870.7\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the perpendicular component of the rate of change of the child's momentum is $870.7\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2$.

At speeds far slower than the speed of light, the net force exerted on the kid equals the rate change of momentum, and the magnitude of the perpendicular rate change is given by

$$\begin{gathered} F_{net}=\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt} \\ =\frac{d\overrightarrow{p}}{dt} \\ =870.7\mathrm{N} \end{gathered}$$

So, the net force acting on the child is $870.7\mathrm{N}$.

The child is subjected to two forces: the cord's tension force $F_T$and its weight $mg$,so the net force of both forces is

$$\begin{gathered} F_{net}=F_T-mg \\ {F}_T=F_{net}+mg \end{gathered}$$

Now put the values for$m$,$F_{net}$,$g$and$R$to get the tension force in the cord

$$\begin{gathered} F_T=F_{net}+mg \\ =870.7\mathrm{N}+\left(26\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =1125.5\mathrm{N} \end{gathered}$$

Thus, the magnitude of the force that the elastic cord exerts on the child is $1125.5N$.

The cord's tension force $F_T$,

$F_T=kx$

Where$k$denotes the cord's stiffness and$x$denotes the cord's stretched length or elongation.

And it's equal to the difference between the spring's original length and its ultimate length after extension, which figured out by

$$\begin{gathered} x=R-L \\ =4.30\mathrm{m}-4.22\mathrm{m} \\ =0.08\mathrm{m} \end{gathered}$$

Now put the values for$F_T$and$x$and solve it for$k$,

$$\begin{gathered} k=\frac{F_T}{x} \\ =\frac{1125.5\mathrm{N}}{0.08\mathrm{m}} \\ =1.4\times {10}^4\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the stiffness of the cord is $1.4\times {10}^4\mathrm{N}/\mathrm{m}$.