Question

a 4500kg roller coaster starts from rest at the top of a 70m tall hill then goes down and back up a 55m tall hill.

a. what is the PE at the top of the 70m tall hill? b. what is the PE at the top of a 55m tall hill? c. what is the KE at the top of a 55m hill? d. what is the velocity at the top of the 55m hill?

Answer 1

here,

the mass of roller coaster , m = 4500 kg

the initial height , h1 = 70 m

the final height , h2 = 55 m

a)

the PE at the top of the hill , PE1 = m * g * h1

PE1 = 4500 * 9.81 * 70 J

PE1 = 3.09 * 10^6 J

b)

the PE at the top of the hill 2, PE2 = m * g * h2

PE2 = 4500 * 9.81 * 50 J

PE2 = 2.21 * 10^6 J

c)

let the kinetic energy at 2 be KE

using conservation of energy

KE + PE2 = PE1

KE + 2.21 * 10^6 J = 3.09 * 10^6 J

KE = 8.83 * 10^5 J

d)

mass , m = 4500 kg

let the velocity at the top of hill be v

KE = 0.5 * m * v^2

8.83 * 10^5 = 0.5 * 4500 * v^2

solving for v

v = 6.26 m/s

the velocity at the top of hill is 6.26 m/s