Page 58 - Physical Chemistry

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39
Work Section 2.1
Suppose a force F acts on a body while the body undergoes an infinitesimal displace- Classical Mechanics
ment dx in the x direction. The infinitesimal amount of work dw done on the body by
the force F is defined as

dw F dx (2.8)*
x
where F is the component of the force in the direction of the displacement. If the
x
infinitesimal displacement has components in all three directions, then
dw F dx F dy F dz (2.9)
z
x
y
Consider now a noninfinitesimal displacement. For simplicity, let the particle be
moving in one dimension. The particle is acted on by a force F(x) whose magnitude
depends on the particle’s position. Since we are using one dimension, F has only one
component and need not be considered a vector. The work w done by F during
displacement of the particle from x to x is the sum of the infinitesimal amounts of
1 2
work (2.8) done during the displacement: w F(x) dx. But this sum of infinitesi-
mal quantities is the definition of the definite integral [Eq. (1.59)], so
x 2
w F1x2 dx (2.10)
x 1
In the special case that F is constant during the displacement, (2.10) becomes

w F1x x 2 for F constant (2.11)
1
2
From (2.8), the units of work are those of force times length. The SI unit of work
is the joule (J):
2
1 J 1 N m 1 kg m >s 2 (2.12)
Power P is defined as the rate at which work is done. If an agent does work dw in
time dt, then P dw/dt. The SI unit of power is the watt (W): 1 W 1 J/s.

Mechanical Energy
We now prove the work–energy theorem. Let F be the total force acting on a particle,
and let the particle move from point 1 to point 2. Integration of (2.9) gives as the total
work done on the particle:
2 2 2
w F dx F dy F dz (2.13)
x y z
1 1 1
Newton’s second law gives F ma m(dv /dt). Also, dv /dt (dv /dx) (dx/dt)
x x x x x
(dv /dx)v . Therefore F mv (dv /dx), with similar equations for F and F . We have
x x x x x y z
F dx mv dv , and (2.13) becomes
x x x
2 2 2
w mv dv mv dv mv dv
x x y y z z
1 1 1
1
2
2
2
2
1
2
2
w m1v v v 2 m1v v v 2 (2.14)
2 x2 y2 z2 2 x1 y1 z1
We now define the kinetic energy K of the particle as
1 2 1 2 2 2
K mv m1v v v 2 (2.15)*
2
2
y
x
z
The right side of (2.14) is the final kinetic energy K minus the initial kinetic energy K :
1
2
w K K ¢K one-particle syst. (2.16)
1
2