p05_026.pdf
(
82 KB
)
Pobierz
Chapter 5 - 5.26
26. We assume the direction of motion is +
x
and assume the refrigerator starts from rest (so that the speed
being discussed is the velocity
v
which results from the process). The only force along the
x
axis is the
x
component of the applied force
F
.
(a) Since
v
0
=0, the combination of Eq. 2-11 and Eq. 5-2 leads simply to
F
x
=
m
v
t
=
⇒
v
i
=
F
cos
θ
i
m
t
for
i
=1 or 2 (where we denote
θ
1
=0and
θ
2
=
θ
for the two cases). Hence, we see that the ratio
v
2
over
v
1
is equal to cos
θ
.
(b) Since
v
0
=0, the combination of Eq. 2-16 and Eq. 5-2 leads to
F
x
=
m
v
2
2∆
x
=
⇒
v
i
=
2
F
cos
θ
i
m
∆
x
for
i
=1 or 2 (again,
θ
1
=0a
nd
θ
2
=
θ
is used for the two cases). In this scenario, we see that the
ratio
v
2
over
v
1
is equal to
√
cos
θ
.
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
p05_007.pdf
(90 KB)
p05_001.pdf
(69 KB)
p05_013.pdf
(77 KB)
p05_006.pdf
(81 KB)
p05_002.pdf
(76 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap06
Zgłoś jeśli
naruszono regulamin