Physical Chemistry Third Edition

388 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

and is the second time derivative of the position vector. If a particle moves in three

dimensions, Newton’s second law is expressed by the vector equation

Fmam

dv

dt

m

d^2 r

dt^2

(Newton’s second law) (9.2-4)

0

0

2.5

5.0

7.5

10.0

12.5

Velocity/m s^21

Position/m

0.5 1.0

Time/s

Figure 9.1 Graph for Example 9.1.
Position and Velocity as a Function of
Time.This figure shows how the velocity
decreases linearly to zero and the coor-
dinate attains a constant value.

EXAMPLE 9.1

A particle moving on a horizontal dry surface is subject to a frictional force in the direction

opposite to its velocity. There is no frictional force if the object is stationary. The magnitude

of the frictional force is assumed to be equal to a coefficient of friction times the weight

(gravitational force) of the object. Find an expression for the position and velocity of an object

of mass 1.000 kg if it moves in thexdirection with an initial velocityvx(0) 10 .0ms−^1

and an initial position ofx0 and if the coefficient of friction is equal to 0.800.

Solution

So long asvxis nonzero, the frictional force is

Fx−(0.800)(1.000 kg)

(

9 .80 m s−^2

)

− 7 .84 kg m s−^2 − 7 .84 N

Newton’s second law gives theequation of motion:

Fxm

d^2 x

dt^2

m

dvx

dt

so that

dvx

dt

−

Fx

m

−

7 .84 kg m s−^2

1 .00 kg

− 7 .84 m s−^2

We integrate this equation fromt0tott′:

∫t′

0

dvx

dt

dt−

(

7 .84 m s−^2

)∫t′

0

dt

vx(t′)−vx(0)−(7.84 m s−^2 )t′

vx(t′) 10 .0ms−^1 −(7.84 m s−^2 )t′

This solution applies only untilvxvanishes, at which time the force stops and the particle

remains stationary. This occurs when

t′

10 .0ms−^1

7 .84ms−^2

 1 .28 s

The position is given by another integration:

x(t′)−x(0)

∫t′

0

dx

dt

dt

∫t′

0

vxdt

∫t′

0

[

10 .0ms−^1

(

7 .84 m s−^2

)

t

]

dt

In this equation we have used the expression for the velocity in whicht′was replaced byt.

The integration yields

x(t′)

(

10 .0ms−^1

)

t′−

1

2

(

7 .84ms−^2

)

t′^2

This solution applies only for values oft′no larger than 1.28 s. The position and velocity are

shown in Figure 9.1.