Question

1) A roller coaster car on the leveled portion of a track is moving at a speed of 35 m/s, heading toward a hump. Based on the principle of conservation of energy, predict what is the velocity of the car on the top of the hump if the lost of energy due to friction and air resistance is negligiable. The highest point of the hump is 14 above the leveled track.

Answer 1

Let us consider a roller coaster of mass m

Let the speed of the roller coaster on the leveled track is V = 35 m/s

Now the kinetic energy of the roller coaster during the motion on the leveled track will be K1 = ½ * m * V²

Let the maximum height reached by roller coaster be h = 14 m

Now the potential energy of the roller coaster is U = m * g * h

Let the velocity at this stage be v

So kinetic energy will be K2 = ½ * m * v²

According to law of conservation of the energy,

Kinetic energy on leveled track is equal to potential energy at maximum height + kinetic energy at that point

So     m * g * h + ½ * m * v² = ½ * m * V²

              ( g * h ) + ( ½ * v² ) = ½ * V²

              ( ½ * V² ) - ( g * h ) = ½ * v²

( ½ * 35 * 35 ) – (9.81 * 14) = ½ * v²

2 *   ( 612.5 – 137.34) = v²

2 * 475.16 = v²

            Finally v² = 950.32

v = 30.83 m/s

so the speed of the coaster at height 14 m is v = 30.83 m/s