03prob_11

03prob_11 - Assignment#3 Page 1 of 2 UC Davis Box  1 Box...

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Unformatted text preview: Assignment #3, 4/15/11 Page 1 of 2 UC Davis Box  1: Box  2: girl: F N1s F N2s F Ngs F 1e F 2e F ge F T1r F F f ge F T2r F √ 1 R √ F √ 2 =R √-F √ 1 135° Box  1: Box  2: girl: F N1s F N2s F Ngs F 1e F 2e F ge F T1r F F F T2r F √ 1 R √ F √ 2 =R √-F √ 1 135° Physics 9A-C Assignment #3 Cole [4.4]

 F x 
=
F 
cos
oe
is
the
angle
that
the
rope
makes
with
the
ramp
(oe
=
30°
in
this
problem),
so

 F x = F cos θ 
õ
 F = F x cos θ = 60 N cos 30 ° 
=
69.3
N

b) 
F y 
=
F 
sinoe
=
 F x 
 tan
oe
=
34.6
N. [4.16]

With 
vø 
=
0,

 v 2 = v 2 + 2 a Δ x 
õ

 a = v 2 2 Δ x = 3 × 10 6 m ( ) 2 2 1.8 × 10 − 2 m ( ) 
=
2.50
ª
10 14 
m/s 2 b)
 v
=
vø
+
at
õ
 
 t = v t = 3 × 10 6 m/s 2.5 × 10 14 m/s 2 
=
1.20
ª
10-2 
s
This
time
is
also
the
distance
divided
by
 the
average
speed.

c)

 F
=
ma
 =
(9.11
ª
10-31 
kg)
(2.50
ª
10 14 
m/s 2 )
=
2.28
ª
10-16 
N. [4.34]

a)
 v
=
0
 where
 v 2 = v 2 − 2 a Δ x ,
and
 v
=
vø
-at.

 Solving
for
 t:
 
 t = v a = 2 Δ x v = 2 0.13 m ( ) 350 m/s 
 =
7.43
ª
10-4 
s
where
 a = v 2 2 Δ x .

b) 
F
=
ma 
=
 m v 2 2 Δ x =
 1.8 × 10 − 3 kg ( ) 350 m/s ( ) 2 2 0.13 m ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 
=
848
N


 Using
 a = Δ v t 
gives
the
same
result....
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This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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03prob_11 - Assignment#3 Page 1 of 2 UC Davis Box  1 Box...

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