A person standing on a bridge at a height of 104 m above a river drops a 0.254 k
ID: 1487526 • Letter: A
Question
A person standing on a bridge at a height of 104 m above a river drops a 0.254 kg rock.
(a) What is the rock's mechanical energy at the time of release relative to the river?
Emech =
(b) What are the rock's kinetic and potential energies after it has fallen 75.0 m?
U =
K =
(c) What are the rock's velocity and mechanical energy just before hitting the water?
v =
Emech =
(d) Answer parts (a)-(c) taking the reference point(h=0) at the elevation where the rock is released.
Emech =
U =
K =
v =
Emech =
Explanation / Answer
a)
Emech = mgh = 0.254*9.8*104 = 258.8768 J
b) at h = (104 - 75) = 29 m
Mechenical Energy (U) = mgh = 0.254*9.8*29 = 72.1868 J
KE ( K) = 258.8768 - 72.1868 = 186.69 J
c) Total Potential energy (mechenical energy) converted in Kinetic energy
KE = PE
0.5*mv2 = mgh
v = sqrt(2*9.8*104) = 45.15 m/s
Emech = zero {Total Mechenical energy converted in Kinetic energy}
d)
if h=0
then all perameters are zero.
Emech = 0
U = 0
K =
Emech = 0
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