A 25.0 kgchild plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are$\mathbf{42}\mathbf{.}\mathbf{0}\mathbf{°}$from the vertical and releases her from rest. (a) What is her potential energy just as she is released compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

The given data can be listed below as,

• The mass of a child is, $m=25.0\mathrm{kg}$.
• The length of support ropes is, $L=2.20\mathrm{m}$.
• The angle up to which her brother pulls the ropes from the vertical is, $\theta =42°$.

Whenever an object moves under the action of an external force, then there would be mechanical work performed by force. The relation between the mechanical work and external force is directly linear.

The expression of the difference in height between the initial and final position of the swing is expressed as,

$$\begin{gathered} ∆h=L-L\cos \theta \\ =L\left(1-\cos \theta \right) \end{gathered}$$

Here, $∆h$ is the difference in height between the initial and the final position of the swing.

Substitute all the known values in the above expression.

$$\begin{gathered} ∆h=\left(2.20\mathrm{m}\right)\left(1-\cos 42°\right) \\ \approx 0.565\mathrm{m} \end{gathered}$$

The relation of the potential energy just as she is released compared with the potential energy at the bottom of the swing’s motion is expressed as,

$PE=mg∆h$

Here, PE is the potential energy just as she is released compared with the potential energy at the bottom of the swing’s motion and g is the gravitational acceleration whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the known values in the above equation.

$$\begin{gathered} PE=\left(25.0\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(0.565\mathrm{m}\right) \\ \approx 1.39\times {10}^2\mathrm{kg}\cdot {\mathrm{m}}^2/{\mathrm{s}}^2 \\ \approx \left(1.39\times {10}^2\mathrm{kg}\cdot {\mathrm{m}}^2/{\mathrm{s}}^2\right)\times \left(\frac{1\mathrm{J}}{1\mathrm{kg}\cdot {\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ \approx 1.39\times {10}^2\mathrm{J} \end{gathered}$$

Thus, the potential energy just as she is released compared with the potential energy at the bottom of the swing’s motion is $1.39\times {10}^2\mathrm{J}$.

The relation of speed of the child at the bottom is expressed as,

$$\begin{gathered} KE=PE \\ \left(\frac{1}{2}m{v}^2\right)=\left(mg∆h\right) \\ v=\sqrt{2g∆h} \end{gathered}$$

Here, KE is the kinetic energy of the child at the bottom, v represents the speed of the child at the bottom.

Substitute all the known values in the above equation.

$$\begin{gathered} v=\sqrt{2\left(9.18\mathrm{m}/{\mathrm{s}}^2\right)\left(0.565\mathrm{m}\right)} \\ =3.33\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the child at the bottom is $3.33\mathrm{m}/\mathrm{s}$.

Now, when swinging the tension in rope would be perpendicular to the distance covered by the child and as such the angle would be$\theta \text{'}=90°$.

The relation of the amount of work does the tension in the ropes do as she swings from the initial position to the bottom of the motion is expressed as,

$W=\left(Td\cos \theta \text{'}\right)$

Here, $W$is the work does the tension in the ropes do as she swings from the initial position to the bottom of the motion, $T$ is the tension in the string, $d$ is the displacement of the child.

Substitute all the known values in the above equation.

$\begin{array}{l}W=\left(T·d\right)(\cos 90°)\\ =\left(T·d\right)\left(0\right)\\ =0\end{array}$

Thus, the amount of work does the tension in the ropes do as she swings from the initial position to the bottom of the motion is 0.