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PHYS Exam #2 Review

# PHYS Exam #2 Review - 2/r or n(mv 2 = mg Chapter 7 Dot...

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IMPORTANT PHYS REVIEW – Exam #2 TIER 408 – 3:15 Thursday Chapter 5 - Friction – 2 questions. Chapter 6 – Centripetal motion – 2 Questions. Chapter 7 – Work/Energy – 2 Questions. Chapter 8 – Conservation of energy and work on a system – 3 questions – 1 work-out problem. Chapter 9 - Impulse – Collision – 3 Questions – 1 work out. Review: Homeworks 5, 6, 7, and 8 NOT 9 – worked out solutions available tomorrow Chapter 5 friction on a plane pulleys on a flat surface m 1 g – T = m 1 a T-u k n 2 = m 2 a m 1 g – T = m 1 a T – m 2 gsintheta-Fk = m 2 a Two blocks attached by a string T 1 costheta – u 1 n 1 – T 2 – m 1 a T 2 – u k n 2 = m 2 a Stopping distance of a car u k n = m car a car . Find acceleration using kinematics Max acceleration of a car with an object on it – what acceleration is needed to move it? Chapter 6 – circular motion F net = (mv 2 )/r T = (mv 2 )/r Swinging cone – Tsin θ = (mv 2 )/r Swinging on a string in two dimensions T – mg = (mv 2 )/r

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Roller coaster Bottom N – mg = (mv 2 )/r Top: N – mg = -(mv
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Unformatted text preview: 2 )/r or n + (mv 2 ) = mg Chapter 7: Dot product H = F Â· d W tot = F net Â· d W tot = K f- K i W = .5k(x i 2 â€“ x f 2 ) Fs = -kx Car moving at 5 m/s brought to stop. U = .2 what is the stopping distance? W f = k f- k i-um gd = -.5m v 2 Chapter 8 Potential = mgy Elastic = .5kx 2 W nc = P + K Ei + Wnc = Ef m 1 g 1 h 1 = m 1 g 1 h f + .5m 1 v 2 m 2 g 2 h 2 = m 2 g 2 h f + .5m 1 v 2 Impulse = Force x time Impulse = âˆ† v x t Bouncing: Init = 5 m/s Final = 3 m/s Change is 8 m/s NOT 2 Imp = P âˆ† M1 is moving at a velocity of 5i â€“ 3j â€“ weighs 2 kg Momentum = 10i â€“ 6j M2 is moving at -7i â€“ 3j weighs 4 kg Momentum = -28i â€“ 12j Total K = -18i â€“ 18j Total weight = 6 So total v = -18/6i, -18/6j Inelastic collision m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v f Elastic collision: P.E. - m 1 v 1 + m 2 v 2 = m 1 V 1f + m 2 v 2f K.E. - Â½m 1 v 1 2 + Â½m 2 v 2 2 = Â½m 1 v 1f 2 + Â½m 2 v 2f 2...
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