Question

A child whose weight is 267 Nslides down a 6.1 mplayground slide that makes an angle of $\mathbf{20}\mathbf{°}$with the horizontal. The coefficient of kinetic friction between slide and child is 0.10. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.457 m/s, what is her speed at the bottom?

Short answer

1. The amount of energy transferred to thermal energy is 153 J
2. If the child starts with a speed of 0.457 m/s, the final speed at the bottom is 5.5 m/s

Step-by-step solution

Listing the given quantities

The weight of the child = 267 N

The length of the slide, L = 6.1 m

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

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

Understanding the concept of conservation of energy

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

Step 3(a): Calculation of amount of energy transferred to 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 $F_f={\mu }_{k}N$

But, according to the diagram,$N=mg\cos \theta$

$$\begin{gathered} F_f={\mu }_{k}mg\cos \theta \\ =0.10\times 267\times \cos 20 \\ =25.1N \end{gathered}$$

Now, the thermal energy can be calculated as,

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

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

Step 4(b): Calculation of speed

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

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

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,

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

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