479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:55 PM Page 503
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R1–34. The 400-kg mine car is hoisted up the incline using
the cable and motor M. For a short time, the force in the
cable is F = (3200t 2 ) N, where t is in seconds. If the car has
an initial velocity v 1 = 2 m>s when t = 0, determine its
velocity when t = 2 s.
v 1 2 m/s
M
17
15
8
+Q©F x¿ = ma x¿ ; 3200t 2 - 400(9.81)a 8
17 b = 400a a = 8t2 - 4.616
dv = adt
y
2
dv = A8t 2 - 4.616Bdt
L2
L 0
v = 14.1 m>s
Ans.
Also,
mv 1 +©
L
F dt = mv 2
2
+Q 400(2) + 3200 t 2 dt - 400(9.81)(2 - 0)a 8
L 17 b = 400v 2
0
800 + 8533.33 - 3693.18 = 400v 2
v 2 = 14.1 m>s
R1–35. The 400-kg mine car is hoisted up the incline using
the cable and motor M. For a short time, the force in the
cable is F = (3200t 2 ) N, where t is in seconds. If the car has
an initial velocity v 1 = 2 m>s at s = 0 and t = 0, determine
the distance it moves up the plane when t = 2 s.
v 1 2 m/s
M
17
15
8
©F x¿ = ma x¿ ; 3200t 2 - 400(9.81)a 8
17 b = 400a a = 8t2 - 4.616
dv = adt
y
t
dv = A8t 2 - 4.616Bdt
L2
L 0
v = ds
dt = 2.667t3 - 4.616t + 2
s
2
ds = A2.667t 3 - 4.616t + 2Bdt
L2
L 0
s = 5.43 m
Ans.
503