Question

3. A 850-kg roller coaster starts from rest 12.0 m above the ground. Position A is at ground level while position B is 8.0 m above the ground. Assume the coaster is frictionless. (a) What is the speed of the coaster at A? (b) What is the speed of the coaster at B?

Answer 1

a) 15.5 m/s

b) 8.96 m/s

Step-by-step explanation:

a) KE (at A) = PE (at top)

`(1)/(2)`
`mv^(2)` = mgh

masses cancel out:

`(1)/(2)`
`v^(2)` = gh

v =
`√(2 gh)`

v =
`√(2 *10*12)`

v ≈ 15.5 m/s

b) KE (at A) = PE (at B) + KE (at B)

(Note: KE at point A will be written as KA, PE at point B will be written as PB, and KE at point B will be written as KB)

KA =
`(1)/(2)`
`mv_(A) ^(2)`

= 0.5 x 850 x
`15.5^(2)`

= 102106.25

PB = mgh

= 850 x 10 x 8

= 68000

KB =
`(1)/(2)`
`mv_(B) ^(2)`

KA = PB + KB

102106.25 = 68000 +
`(1)/(2)`
`mv_(B) ^(2)`

`(1)/(2)`
`mv_(B) ^(2)` = 34106.25

m
`v_(B) ^(2)` = 68212.5

`v_(B) ^(2)` = 80.25

`v_(B)` ≈ 8.96 m/s