Question

A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. as the coaster moves from its initial height to its lowest position, 1700j of energy is transformed to thermal energy by friction. in order for the roller coaster to safely travel over the second hill, it must be moving at a velocity of 4.6m/s or less at the top of the second hill. what is the maximum height the second hill can be

Answer 1

Answer:

10.6 meters.

Explanation:

We use the law of conservation of energy, which says that the total energy of the system must remain constant, namely:

$\frac{1}{2}mv_i^2+mgh_i-1700j=\frac{1}{2}mv_f^2+mgh_f$

In words this means that the initial kinetic energy of the roller coaster plus its gravitational potential energy minus the energy lost due to friction (1700j) must equal to the final kinetic energy at top of the second hill.

Now let us put in the numerical values in the above equation.

$m=100kg$

$h_i=10m$

$v_i= 6m/s$

$v_f=4,6m/s$

and solve for $h_f$

$h_f= \frac{\frac{1}{2}mv_i^2+mgh_i-1700j-\frac{1}{2}mv_f^2}{mg} =\boxed{ 10.6\:meters}$

Notice that this height is greater than the initial height the roller coaster started with because the initial kinetic energy it had.