91Ó°ÊÓ

The generated thermal energy in a system is equal to the loss of mechanical energy from the system.

Given data:

The mass of the child is \(m = 16.0\;{\rm{kg}}\).

The height of the slide is \(h = 2.20\;{\rm{m}}\).

The initial speed of the child at the height is \({v_o} = 0\) as it starts from rest.

The speed of the child at the bottom is \(v = 1.25\;\frac{{\rm{m}}}{{\rm{s}}}\).

Let \(\Delta H\) be the generated thermal energy in this process.

The total mechanical energy of the system is calculated as follows:

\(\begin{aligned}{E_1} &= mgh + \frac{1}{2}mv_o^2\\ &= mgh + \left( {\frac{1}{2}m \times {0^2}} \right)\\ &= mgh\end{aligned}\)

The total mechanical energy of the system at the bottom is calculated as follows:

\(\begin{aligned}{E_2} &= \left( {mg \times 0} \right) + \frac{1}{2}m{v^2}\\ &= \frac{1}{2}m{v^2}\end{aligned}\)

Now the thermal energy is equal to the loss in mechanical energy. Then,

Now the thermal energy is equal to the loss in mechanical energy. Then,

\(\begin{aligned}\Delta H &= {E_1} - {E_2}\\ &= mgh - \frac{1}{2}m{v^2}\\ &= \left[ {\left( {16.0\;{\rm{kg}}} \right) \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times 2.20\;{\rm{m}}} \right] - \left[ {\frac{1}{2} \times \left( {16.0\;{\rm{kg}}} \right) \times {{\left( {1.25\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}} \right]\\ &= 332.46\;{\rm{J}}\end{aligned}\)

Hence, the generated thermal energy in this process is 332.46 J.