Question

[Use g = 10 m/s2]

A single roller coaster car is at rest at the top of a hill of height h0 = 613 m. The cart rolls down over the track through points A, B, and C (with heights shown below). Find the coaster's speed at each of these points.

hA = 8 m
hB = 368 m
hC = 568 m

Speed at point A =

Speed at point B =

Speed at point C =

Answer 1

Answer: Kinetic energy: 1/2 * m * v^2

Explanation: Where:

h0=337m

hA = 17 m

hB = 157 m

hC = 317 m

Initial energy = final energy

PE0 + KE0 = PE2 + KE2

KE0 = velocity = 0

KE= 1/2 mv^2 = 0

PE0 = initial potential energy = mgh

m=mass

h=height

g= gravity= 10 m/s^2

For point A.

Instead oh h0=337 and hA=17m we can state that

H0=337-17 = 320

HA= 0 (so PEA=0 )

PE0 = KEA

mgh0 =1/2 m va^2

gh0 = 1/2 va^2

10 * 320 = 1/2 va^2

√(10 * 320)*2 = va

va = 80 m/s

Same process for point B

mgh0 = 1/2 m vb

g h0 = 1/2 vb

10 * 180 = 1/2 vb^2

√(10 * 180)*2 = vb

vb= 60 m/s

Point C:

H0=337-317=20

vc= √(10 * 20)*2

vc= 20 m/S