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\(20.64\;\frac{{\rm{m}}}{{\rm{s}}}\).
