3. A 0.5-lb slider moves along a semicircular wire in a horizontal plane as shown. The undeformed length of the spring is 8 in., and friction may be neglected. If the slider is released from rest at position A, determine a. The velocity of the slider at position B. b. The force exerted on the slider by the wire at position B. Use Conservation of Energy and Equation of Motion then Newton's law. 12 in 051 12 in. =0.75 IM (a) TA+VA = TB + VB 0+1(0.75) (금)( )°= (0.75) (음)( ) ft/s V%3D (b) EF= ma = m - (0.75) ()( )+N = 05 1 N =(

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Chapter13: Vibrations And Waves
Section: Chapter Questions
Problem 32P: A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is...
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3. A 0.5-lb slider moves along a semicircular wire in a horizontal plane as shown. The
undeformed length of the spring is 8 in., and friction may be neglected. If the slider is
released from rest at position A, determine
a. The velocity of the slider at position B.
b. The force exerted on the slider by the wire at position B.
Use Conservation of Energy and Equation of Motion then Newton's law.
12 in.
0.5 lb
12 in.
=0.75lMA
(a)
TA + VA = TB + VB
0 + }(0.75)()(
)2 =D1(0.75) (금)( )2+1002
v = V ) ~ (
(b)
EF = ma = m
- (0.75)()(
) ft/s
)+N = °
0.5 (
1
N (
lb
Transcribed Image Text:3. A 0.5-lb slider moves along a semicircular wire in a horizontal plane as shown. The undeformed length of the spring is 8 in., and friction may be neglected. If the slider is released from rest at position A, determine a. The velocity of the slider at position B. b. The force exerted on the slider by the wire at position B. Use Conservation of Energy and Equation of Motion then Newton's law. 12 in. 0.5 lb 12 in. =0.75lMA (a) TA + VA = TB + VB 0 + }(0.75)()( )2 =D1(0.75) (금)( )2+1002 v = V ) ~ ( (b) EF = ma = m - (0.75)()( ) ft/s )+N = ° 0.5 ( 1 N ( lb
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