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Chapter 5 - 5.40
40. Referring to Fig. 5-10(c) is helpful. In this case, viewing the man-rope-sandbag as a system means that
we should be careful to choose a consistent positive direction of motion (though there are other ways
to proceed – say, starting with individual application of Newton’s law to each mass). We take
down
as positive for the man’s motion and
up
as positive for the sandbag’s motion and, without ambiguity,
denote their acceleration as
a
. The net force on the system is the difference between the weight of the
man and that of the sandbag. The system mass is
m
sys
= 85 + 65 = 150 kg. Thus, Eq. 5-1 leads to
(85)(9
.
8)
−
(65)(9
.
8) =
m
sys
a
which yields
a
=1
.
3m/s
2
. Since the system starts from rest, Eq. 2-16 determines the speed (after
traveling ∆
y
= 10 m) as follows:
v
=
2
a
∆
y
=
2(1
.
3)(10) = 5
.
1m
/
s
.
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