91Ó°ÊÓ

The weight of the child = 267 N

The length of the slide, L = 6.1 m

The coefficient of friction,${\mathrm{μ}}_k=0.10$

The angle made by slide with the ground$={20}^o$

As the child starts sliding, the friction with the slide surface converts some of the initial energy to thermal energy. The total energy of the child-slide system is conserved during motion. When the child starts with some initial speed, its energy converts to kinetic energy at the bottom with some energy lost as thermal energy.

Formula:

$K.E=\frac{1}{2}m{v}^2$

$P.E=mgh$

$E_{th}=F_{f}x$

Total mechanical energy at the bottom = Total mechanical energy at the top - Thermal energy

The increase in thermal energy of the child-slide system occurs because of the friction between the child and the slide surface. Hence, first, we determine the frictional force as.

The frictional force is given as $F_f={\mathrm{μ}}_{k}N$

But, according to the diagram,$N=mg\cos \mathrm{θ}$

$$\begin{gathered} F_f={\mathrm{μ}}_{k}mg\cos \mathrm{θ} \\ =0.10×267×\cos 20 \\ =25.1N \end{gathered}$$

Now, the thermal energy can be calculated as,

$$\begin{gathered} E_{th}=F_{f}L \\ =25.1×6.1 \\ =153J \end{gathered}$$

Hence, the amount of energy transferred to thermal energy is 153 J

Now,to determine the mass of the child and the height of the slide as follows…

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{267}{9.8} \\ =27.2kg \end{gathered}$$

And from the diagram we can see

$$\begin{gathered} \sin \mathrm{θ}=\frac{h}{L} \\ h=L\sin \mathrm{θ} \\ =6.1\sin 20 \\ =2.1m \end{gathered}$$

The child – slide system follows conservation of energy. Hence we write, ${\left(Totalmechanicalenergy\right)}_{bottom}={\left(Totalmechanicalenergy\right)}_{top}-thermalenergy$$$\begin{gathered} {\left(KE+PE\right)}_{bottom}={\left(KE+PE\right)}_{top}-E_{th} \\ {\left(\frac{1}{2}m{v}^2+mgh\right)}_{bottom}={\left(\frac{1}{2}m{v}^2+mgh\right)}_{top}-{\left(mgh\right)}_{bottom}-E_{th} \\ {\left(\frac{1}{2}m{v}^2\right)}_{bottom}={\left(\frac{1}{2}m{v}^2+mgh\right)}_{top}-{\left(mgh\right)}_{bottom}-E_{th} \end{gathered}$$

At the bottom, h = 0 so the equation simplifies to,

$$\begin{gathered} {\left(\frac{1}{2}m{v}^2\right)}_{bottom}={\left(\frac{1}{2}×27.2×0.{457}^2+267×2.1\right)}_{top}-153 \\ {\left(\frac{1}{2}m{v}^2\right)}_{bottom}=2.86+561-153 \\ {\left(\frac{1}{2}m{v}^2\right)}_{bottom}=411J \\ {\left(v^2\right)}_{bottom}=\frac{411×2}{27.2} \\ {\left(v^2\right)}_{bottom}=30.2 \\ v=5.5m/s \end{gathered}$$

Hence, the final speed at the bottom is 5.5 m/s, if the child starts with a speed of 0.457 m/s,