(II) Suppose the roller-coaster car in Fig. 6–41 passes point 1 with a speed of 1.30 m/s. If the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? The distance traveled is 45.0 m.

The work done by the friction force is equal to the loss in mechanical energy between two points.

Given data:

The height of point 1 is\({h_1} = 32\;{\rm{m}}\).

The height of point 2 is\({h_2} = 0\).

The speed of the roller-coaster when it passes point 1 is\({v_1} = 1.30\;\frac{{\rm{m}}}{{\rm{s}}}\).

The distance between point 1 and point 2 is\(d = 45.0\;{\rm{m}}\).

The friction force is equal to 0.23 times of weight.

Assumptions:

Let the mass of the roller coaster be\(m\).

The friction force on the roller-coaster is\(f = 0.23mg\).

The mechanical energy at point 1is calculated as follows:

\({E_1} = mg{h_1} + \frac{1}{2}mv_1^2\)

The mechanical energy at point 2 is calculated as follows:

\(\begin{aligned}{E_2} &= mg{h_2} + \frac{1}{2}mv_2^2\\ &= \left( {mg \times 0} \right) + \frac{1}{2}mv_2^2\\ &= \frac{1}{2}mv_2^2\end{aligned}\)

The work done against the friction force is\(W = fd\).

Now from energy conservation, you can write,

\(\begin{aligned}W &= {E_1} - {E_2}\\fd &= mg{h_1} + \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2\\0.23mgd &= mg{h_1} + \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2\\0.46gd &= 2g{h_1} + v_1^2 - v_2^2\end{aligned}\)

After further calculation, you will get,

\(\begin{aligned}v_2^2 &= g{h_1} + v_1^2 - 0.46gd\\v_2^2 &= \left[ {2 \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {32\;{\rm{m}}} \right)} \right] + {\left( {1.30\;\frac{{\rm{m}}}{{\rm{s}}}} \right)^2} - \left[ {0.46 \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {45.0\;{\rm{m}}} \right)} \right]\\{v_2} &= 20.64\;\frac{{\rm{m}}}{{\rm{s}}}\end{aligned}\)

Hence, the roller-coaster will reach point 2 with a speed of\(20.64\;\frac{{\rm{m}}}{{\rm{s}}}\).