(II) A roller-coaster car shown in Fig. 6–41 is pulled up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4.

FIGURE 6–41 Problem 40

The roller coaster car is moving up and down the hill. During the motion, the total energy of the car remains constant.

At the top, the kinetic energy is zero, but it has some potential energy. At the bottom, the potential energy is zero, but it has some kinetic energy.

Equate the sum of the kinetic and potential energy values at the initial point and final point. With the help of this, evaluate the speed of the roller coaster car at the required points.

The given data can be listed below as:

• At top point (1), the value of height is \({h_1} = 32{\rm{ m}}\).
• Thespeed of the car at point (1) is\({v_1} = 0{\rm{ m/s}}\).
• At bottom point (2), the value of height is \({h_2} = 0{\rm{ m}}\).
• At top point (3), the value of height is \({h_3} = 26{\rm{ m}}\).
• The speed of the car at point (3) is \({v_3} = 0{\rm{ m/s}}\).
• At top point (4), the value of height is \({h_4} = 14{\rm{ m}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

From the conservation of energy principle, the energies can be expressed as:

\(\begin{aligned}{E_1} &= {E_2}\\K.{E_1} + P.{E_1} + &= K.{E_2} + P.{E_2}\\\frac{1}{2}mv_1^2 + mg{h_1} &= \frac{1}{2}mv_2^2 + mg{h_2}\end{aligned}\)

Here, \(K.{E_1}\) is the kinetic energy at point (1),which is equal to zero, \(P.{E_1}\) is the potential energy of the roller coaster car at point (1), \(K.{E_2}\) is the kinetic energy of the carat point (2),\(P.{E_2}\) is the potential energy at point (2) whose value is zero, and \({v_2}\) is the speed of the car at point (2).

Substitute the values in the above expression.

\(\begin{aligned}0 + mg{h_1} &= \frac{1}{2}mv_2^2 + 0\\{v_2} &= \sqrt {2g{h_1}} \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{v_2} &= \sqrt {2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 32{\rm{ m}}} \\ &= 25.06{\rm{ m/s}}\end{aligned}\)

Thus, the speed at point (2) is 25.06 m/s.

From the conservation of energy principle, the energies can be expressed as:

\(\begin{aligned}{E_1} &= {E_3}\\K.{E_1} + P.{E_1} + &= K.{E_3} + P.{E_3}\\\frac{1}{2}mv_1^2 + mg{h_1} &= \frac{1}{2}mv_3^2 + mg{h_3}\end{aligned}\)

Here, \(K.{E_3}\) is the kinetic energy of the car at point (3),\(P.{E_3}\) is the potential energy at point (3), and \({v_3}\) is the speed of the car at point (3).

Substitute the values in the above expression.

\(\begin{aligned}0 + mg{h_1} &= \frac{1}{2}mv_3^2 + mg{h_3}\\{v_3} &= \sqrt {2g\left( {{h_1} - {h_3}} \right)} \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{v_3} &= \sqrt {2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {32{\rm{ m}} - 26{\rm{ m}}} \right)} \\ &= 10.85{\rm{ m/s}}\end{aligned}\)

Thus, the speed at point (3) is 10.85 m/s.

From the conservation of energy principle, the energies can be expressed as:

\(\begin{aligned}{E_1} &= {E_4}\\K.{E_1} + P.{E_1} + &= K.{E_4} + P.{E_4}\\\frac{1}{2}mv_1^2 + mg{h_1} &= \frac{1}{2}mv_4^2 + mg{h_4}\end{aligned}\)

Here, \(K.{E_4}\) is the kinetic energy of the carat point (4),\(P.{E_4}\) is the potential energy at point (4), and \({v_4}\) is the speed of the car at point (4).

Substitute the values in the above expression.

\(\begin{aligned}0 + mg{h_1} &= \frac{1}{2}mv_4^2 + mg{h_4}\\{v_4} &= \sqrt {2g\left( {{h_1} - {h_4}} \right)} \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{v_4} &= \sqrt {2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times \left( {32{\rm{ m}} - 14{\rm{ m}}} \right)} \\ &= 18.79{\rm{ m/s}}\end{aligned}\)

Thus, the speed at point (4) is 18.79 m/s.