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Unformatted text preview: 42. (a) Using Eq. 2328, we ﬁnd
F =
= (8.00 × 10−5 C)(3.00 × 103 N/C)ˆ + (8.00 × 10−5 C)(−600 N/C)ˆ
i
j
(0.240 N)ˆ − (0.0480 N)ˆ .
i
j Therefore, the force has magnitude equal to
F= (0.240 N)2 + (0.0480 N)2 = 0.245 N , and makes an angle θ (which, if negative, means clockwise) measured from the +x axis, where
θ = tan−1 Fy
Fx = tan−1 −0.0480 N
0.240 N = −11.3◦ . (b) With m = 0.0100 kg, the coordinates (x, y ) at t = 3.00 s are found by combining Newton’s second
law with the kinematics equations of Chapters 24:
x=
y = 1
Fx t 2
(0.240)(3.00)2
ax t 2 =
=
= 108 m ,
2
2m
2(0.0100)
Fy t 2
(−0.0480)(3.00)2
12
ay t =
=
= −21.6 m .
2
2m
2(0.0100) ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Force

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