Biomedical Engineering Reference
In-Depth Information
EXAMPLE 5.10
A person weighing 70 kg is skiing from the top of a hill that is 50m tall and has a slope of 1
in 20 measured along the slope. He starts at a speed of 1 m/s, after which he skies down to
the bottom of the hill, where his speed increased to 25 m/s. How much energy has he lost?
Solution:
2
2
2
2
1[m/s]
25 [m/s]
+
50[m]
=
+
0
+
h
w
2
2
2 * 9.81[m/s ]
2 * 9.81[m/s ]
18.196 m
[]
h
==
w
2
70[kg] * 9.81[m/s ] * 18.196[m]
Energy lost =
mgh
12, 495J
=
=
w
A simplified version of conservation of energy principle is the conservation of
mechanical energy where the frictional loss is neglected. The sum of the kinetic
energy,
Δ
E
KE
, and the potential energy,
Δ
E
PE
, is referred as the total mechanical
E
ME
. If the system is isolated, with only conservative forces acting on
its parts, the total mechanical energy of the system is a constant. The mechanical
energy can be transformed between its kinetic and potential forms, but cannot be
destroyed. Rather, the energy is transmitted from one form to another. Any change
in the kinetic energy will cause a corresponding change in the potential energy and
vice versa. For objects in rotary motion, kinetic energy is calculated using angular
velocity. Potential energy is a form of stored energy and is a consequence of the
work done by a force. Examples of forces that are associated with potential energy
are the gravitational and the electromagnetic fields and, in biomechanics, a spring.
For a body moving under the influence of a force
F
, the change in potential energy
is given by
energy,
Δ
2
∫
Δ=−
E
F ds
(5.41)
PE
1
where 1 and 2
represent the initial and final positions of the body, respectively.
Equation (5.41) states that the performance of work requires the expenditure of
energy. This relation is commonly referred to as the
work-energy relation
and can
be applied to kinetic energy as well.
f
∫
Δ=
EF ds
KE
i
Consider a body of mass
m
being accelerated by a compressed spring. The
compressive force exerted by a spring within the linear range is given by (5.15),
f
1
2
(
)
∫
Δ=−
««««
=
2
−
2
PE
2
1
i
