As always, in applying Newton’s second law of motion, we start by drawing a free body
diagram:
m
F
a
In order to keep track of the vector nature of the quantities involved we apply Newton’s 2nd Law
in vector form (equation 14-1):
= 1
a
∑F
m
In the case at hand the sum of the forces is just the one force F , so:
1
a =
F
m
Solving for F, we arrive at:
F = a
m
multiplying both sides by ∆t we obtain
F∆t = ma t
∆
Given that the force is constant, the resulting acceleration is constant. In the case of a constant
acceleration, the acceleration can be written as the ratio of the change in v
that occurs during the
time interval ∆t, to the time interval ∆t itself.
v
= ∆
a
t
∆
Substituting this into the preceding expression yields:
v
∆
∆
F t = m
t
∆
∆t
F∆t = m v
∆
The change in velocity can be expressed as the final velocity v ′ (the velocity at the end of the
time interval during which the force acts) minus the initial velocityv (the velocity at the start of
the time interval): v
∆ = v ′−v . Substituting this into F∆t = m v
∆ yields
F∆t = (
m v ′− v )
which can be written as
F∆t = mv ′− mv
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