Question

6. A 200-kg roller coaster reaches the top of the steepest hill with a speed of 5.80 km/h. It then descends the hill, which is at an angle of 35° and is 41.0 m long. What will its kinetic energy be when it reaches the bottom? Assume µk = 0.18.

Answer 1

The two forces that are of importance are the force of gravity and the force of friction. Gravity is accelerating the roller coaster down the hill. Friction acts counter to gravity. Because the roller coaster is on this incline, the acceleration (a) due to gravity is equal to:
=g * sin 35

The force of Friction (F_f) is equal to:

=µ_k *N

The normal force (N) is equal to mg on a horizontal surface. On this incline it is equal to:

N= m * g * cos 35

The difference between the force of gravity (m * g * sin35) and the Frictional force
(µ_k *mg*cos35) will equal the net force acting on the roller coaster.

F = m * a = Force of gravity - Frictional force

m * a = (m * g * sin35) - (µ_k *m*g cos35)

m cancels out, so:

a = (g *sin35) - (µ_k *g cos35)
a=4.18 m/s²

Now convert the initial velocity into m/s:

5.8km/h x ( 1000m/km) x (1 hr/3600sec)

= 1.61 m/s

We know the distance (d = 41m), the initial velocity (V_o = 1.61 m/s), and the acceleration
(a =4.18 m/s² ). Solve for the final Velocity (V) using this equation:

V² = V_o² + (2 * a * d)

V = [ V² + (2 * a * d)]^0.5

V = [ (1.61)² + (2 * 4.18 x 41)]^0.5

= 18.5896 m/s

So the kinetic energy will be,

KE=0.5mv²

KE=0.5*200*18.5896²

KE=34535.21 J