91Ó°ÊÓ

The given data can be listed below,

• Mass of the roller coaster is 350 kg
• The height at point A is 25.0 m
• The height at point B is 12.0 m

According to conservation law, a conserved quantity for a system that goes through a process is constant both before and after the process.

The roller coaster is at rest at the initial position. So, the initial kinetic energy $K_i$is zero.

The gravitational potential energy is expressed as,

$U_{grav}=mgy$

The gravitational potential energy at point A is obtained by substituting the given values for A, in the above equation,

$$\begin{gathered} U_A=\left(350\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(25.0\mathrm{m}\right) \\ =85750\mathrm{J} \end{gathered}$$

Similarly, the gravitational potential energy at point B is obtained by substituting the given values for B, in the above equation,

$$\begin{gathered} U_B=\left(350\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(12.0\mathrm{m}\right) \\ =41160\mathrm{J} \end{gathered}$$

According to law of conservation of energy, the decrease in gravitational potential energy should be equal to increase in kinetic energy. As the kinetic energy is zero at A, so equation becomes-

$K=U_A-U_B$

For $U_A=85750\mathrm{J}$and $U_B=41160\mathrm{J}$, the kinetic energy of the roller coaster at B is given as-

$$\begin{gathered} K=85750\mathrm{J}-41160\mathrm{J} \\ =44590\mathrm{J} \end{gathered}$$

Now, the final kinetic energy is expressed as,

$K=\frac{1}{2}m{v}^2$

For, $K=44590\mathrm{J}$and $m=350\mathrm{kg}$, the velocity of the roller coaster at B is -

$$\begin{gathered} 44590=\frac{1}{2}\left(350\mathrm{kg}\right)\left({\mathrm{v}}^2\right) \\ \mathrm{v}=\sqrt{\frac{2×44590\mathrm{J}}{350\mathrm{kg}}} \\ \mathrm{v}=15.96\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the roller coaster is 15.96 m/s .

The force with which the tracks are pressed at the point B should be equal to the normal reaction force $\left(n\right)$by the tracks, if is the centrifugal acceleration of the roller coaster and g is the acceleration due to gravity of earth, then-

$$\begin{gathered} ma=n+mg \\ n=m\left(a-g\right) \end{gathered}$$

Substitute the value of acceleration in the above equation.

$$\begin{gathered} n=m\left(\frac{v^2}{R}-g\right) \\ =350\mathrm{kg}\left(\frac{{\left(15.96\mathrm{m}/{\mathrm{s}}^2\right)}^2}{6\mathrm{m}}-9.6\mathrm{m}/{\mathrm{s}}^2\right) \\ =11433.3\mathrm{N} \end{gathered}$$

Thus, the force exerted on the track at point B is 11433.3 N