The total height of Yosemite Falls is 2425 ft. Part A How many more joules of gr

Question

The total height of Yosemite Falls is 2425 ft.

Part A

How many more joules of gravitational potential energy are there for each kilogram of water at the top of this waterfall compared with each kilogram of water at the foot of the falls?

Part B

Find the kinetic energy and speed of each kilogram of water as it reaches the base of the waterfall, assuming that there are no losses due to friction with the air or rocks and that the mass of water had negligible vertical speed at the top.

Part C

Part D

How fast (in m/s and mph) would a 63.0 kg person have to run to have that much kinetic energy?

__________

Part E

Part F

How high (in m and ft) would Yosemite Falls have to be so that each kilogram of water at the base had twice the kinetic energy you found in part B?

Part G

Part H

How high (in m and ft) would Yosemite Falls have to be so that each kilogram of water at the base had twice the speed you found in part C?

Part I

Explanation / Answer

A)

m = mass = 1 kg

h = height = 2425 ft = 739.14 m

Gravitational Potential energy is given as

GPE = mgh = 1 x 9.8 x 739.14

GPE = 7243.6 J

B)

using conservation of energy

Kinetic energy at the base = Gravitational PE at top = 7243.6 J

part c)

m = 1 kg

kinetic energy is given as

KE = (0.5) m v2

7243.6 = (0.5) (1) v2

v = 85.11 m/s

d)

M = mass of person = 63 kg

V = speed of person

kinetic energy of person = kinetic energy of water at base

(0.5) M V2 = 7243.6

(0.5) (63) V2 = 7243.6

V = 15.2 m/s

E)

V = 15.2 m/s = 15.2 (m/s) (1 mile/1609.34 m ) (3600 sec/h ) = 34 mph

F)

Potential energy = 2 x KE at base

mgh = 2 x 7243.6

1 x 9.8 h = 2 x 7243.6

h = 1478.3 m

G)

h = 1478.3 m = 1478.3 m (3.28 ft/ 1 m) = 4848.82 ft

H)

v' = speed at base = 2 x 85.11 = 170.22 m/s

KE' = kinetic energy at base

PE at Top = KE at base

mgh' = KE'

mgh' = (0.5) m v'2

9.8 h' = (0.5) (170.22)2

h' = 1478.3 m

I)

h' = 4850.1 ft

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