30. The stopping force
G
F
and the path of the car are horizontal. Thus, the weight of the
car contributes only (via Eq. 512) to information about its mass (
m
=
W
/
g
= 1327 kg).
Our +
x
axis is in the direction of the car’s velocity, so that its acceleration
(‘‘deceleration”) is negativevalued and the stopping force is in the –
x
direction:
ˆ
i
FF
=−
G
.
(a) We use Eq. 216 and SI units (noting that
v
= 0 and
v
0
= 40(1000/3600) = 11.1 m/s).
()
2
2
22
0
0
(11.1 m/s)
2
1
5
m
v
vv
a
x
a
x
=+Δ
¡
= −
Δ
which yields
a
= – 4.12 m/s
2
. Assuming there are no significant horizontal forces other
than the stopping force, Eq. 51 leads to
G
G
Fm
a
F
=
¡
−=
−
1327
412
kg
m s
2
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 Spring '08
 Any
 Physics, Acceleration, Force, Mass

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