Homework 9
Slides:
5)
E
K
= ½(1000kg)(10m/s)
2
=
50,000J
, E
f
K
 E
i
K
= ½(1000kg)(1600m
2
/s
2
)  ½(1000kg)(900m
2
/s
2
) = (800,000 –
450,000)J =
350,000J
,
With traffic more less energy on 95 because 95 would have less traffic and less stop and go. Without
traffic US1 would consume less energy because the speed would be less.
Acceleration 1 = (10m/s – 0)/(10s) = 1m/s
2
; Force 1 = (1000kg)(1m/s
2
) = 1000N; Distance 1 =
𝜐
o
𝑡
+ ½
𝘢
t
2
=
½ (1m/s
2
)(100 s
2
) = 50m; W = (1000N)(50m) =
50,000J
, P = E/
𝑡
= (50,000J)/10s =
5,000W
Acceleration 2 = (40m/s – 30m/s)/(10s) = 1m/s
2
; Force 2 = (1000kg)(1m/s
2
) = 1000N; Distance 2 =
𝜐
o
𝑡
+
½
𝘢
t
2
= (30m/s)(10s) + ½(1m/s
2
)(100s
2
) = 350m; W = (1000N)(350m) =
350,000J
, P = E/
𝑡
= (350,000J)/10s
=
35,000W
6)
Velocity 1 =
√
(
𝜐
y
2
+
𝜐
x
2
);
𝜐
y1
=
𝜐
yo
+
𝘨
t, t = d/
𝜐
x
= (10m)/(15m/s) = (2/3)s;
𝜐
y1
=
(20m/s)
+
(
10m/s
2
)(2/3)s = (60/3)m/s – (20/3)m/s = (40/3)m/s;
𝜐
=
√
[(40/3)
2
+
(45/3
)
2
]m/s =
(
5/3
√
145
)m/s
; E
P
=
𝑚
g(h+
𝛥
h);
𝛥
h =
𝜐
o
𝑡
+ ½
𝘢
t
2
= (20m/s)(6/9)s + ½(10m/s
2
)(4/9)s
2
= (100/9)m; E
P
=
(5kg)(10m/s
2
)[(100+225)/9]m =
1805.5J
, E
K
= ½
𝑚𝜐
2
= ½(5kg)(3625/9)m
2
/s
2
=
1007.0J
, W
𝘨
=
𝛥
h(
𝑚𝘨
) =
(100/9)(5kg)(10m/s
2
) = (
5000/9)J
Velocity 2 =
𝜐
x
=
15m/s
;
E
P
= mg(h+
𝛥
h);
𝛥
h = (
𝜐
f
2
+
𝜐
o
2
)/2
𝘨
= [(0m/s)
2
+ (20m/s)
2
]/2(10m/s
2
) = 20m; h =
25m +
𝛥
h = 45m; E
P
= (5kg)(10m/s
2
)(45m) =
2250J
, E
K
= ½
𝑚𝜐
2
= ½(5kg)(15m/s)
2
=
562.5J
, W
𝘨
=
𝛥
h(
𝑚𝘨
) =
(20m)(5kg)(10m/s
2
) =
2250J
Velocity 3 =
√
(
𝜐
y
2
+
𝜐
x
2
);
𝜐
y1
=
𝜐
yo
+
𝘨
t, t = d/
𝜐
x
– t
2
= (60m)/(15m/s)  (30m)/(15m/s)= 2s;
𝜐
y1
=
(0m/s)
+
(10m/s
2
)(2s) = 20m/s
;
𝜐
=
√
[
(20m/s)
2
+
(15m/s)
2
] =
√
[
625
]m/s = 25m/s; E
P
=
mg(h
max
+
𝛥
h);
𝛥
h =
𝜐
o
𝑡
+ ½
𝘢
t
2
= (0m/s)(2s) + ½(10m/s
2
)(4s
2
) = 20m; E
P
= (5kg)(10m/s
2
)(4520)m =
1250J
,
E
K
= ½
𝑚𝜐
2
= ½(5kg)(25m/s)
2
=
1562.5J
, W
𝘨
= [(h
max
+
𝛥
h) h
0
](
𝑚𝘨
) = [25(4520)]m(5kg)(10m/s
2
) =
0J
Velocity 4 =
√
(
𝜐
y
2
+
𝜐
x
2
);
𝜐
y1
=
𝜐
yo
+
𝘨
t, t = d/
𝜐
x
– t
2
= (70m)/(15m/s)  (30m)/(15m/s)= (8/3)s;
𝜐
y1
=
(0m/s)
+
(10m/s
2
)(8/3)s = (80/3)m/s;
𝜐
= [
√
(
(80/3)
2
+
(45/3
)
2
]m/s =
√
(8425/9)m/s
; E
P
=
mg(h
max
+
𝛥
h);
𝛥
h =
𝜐
o
𝑡
+ ½
𝘢
t
2
= (0m/s)(8/3s) + ½(10m/s
2
)(64/9)s
2
= (320/9)m; E
P
= (5kg)(10m/s
2
)(45
320/9)m =
472.2J
, E
K
= ½
𝑚𝜐
2
= ½(5kg)(8425/9)m
2
/s
2
=
2340.3J
, W
𝘨
= [(h
max
+
𝛥
h) h
0
](
𝑚𝘨
) = [(225/9)
((405/9)+(320/9))]m(5kg)(10m/s
2
) = (140/9)m(5kg)(10m/s
2
) =
777.8J
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Velocity 5 =
√
(
𝜐
y
2
+
𝜐
x
2
) =
𝜐
y1
=
𝜐
yo
+
𝘨
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 Summer '11
 Ramos
 Electron, Energy, Work, Electric charge, 10m/s, 0.09m, Cosθmax, 2.6J

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