Physics 101 Exam 2

Physics 101 Exam 2 - f,astronaut p p (75kg 25kg)v...

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The applied force is transmitted throughout the rope (remember we talked about pulley arrangements being used as “force multipliers?) The mass has 2 upward forces acting on it and its own weight acting downward. 2 2 y m s m s F ma 2F mg ma 2(180N) (30kg)(9.81 ) a 30kg a 2.19 = - = - = = 1969 Corvette Stingray. mg a c N
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2 y y 2 2 2 m s m s F ma mv N mg Why the negative sign on the right? Because the r acceleration (centripetal) is pointing DOWN, which is in the -y direction! (10 ) v N=m g+ (60kg) 9.8 388N r 30m = - = - = - = ( ) ( ) 2 2 2 m 1 s f f f 2 2 2 2 1 m i i i 2 s 365.6 mv K v 0.32 32% K mv v 644.9 = = = = = 1600kg 2 2 2 2 m m m f i 2 s s 2 ((40 ) (15 ) ) (v v ) W K 1hp P 369hp t t t 4.00s 746W - - Δ = = = = = Use conservation of energy and the information given in the first part to get k.
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i2 f 2 i2 i2 f 2 f 2 2 2 1 1 2 2 2 2 2 i1 i1 f1 f1 2 2 1 1 2 2 2 2 2 2 2 2 E E U K U K k(d ) 0 0 (4m)(2v) 16mv d Now we find k using the first set of data: k U K U K k(d) 0 0 (m)(v) mv k d 16mv d d 4d mv = + = + + = + = + = + + = + = = = Use conservation of momentum: i f f,astronaut
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Unformatted text preview: f,astronaut p p (75kg 25kg)v (0.01kg)(100.0m/s) v 0.01m/s = = + + =-! ! Recoil m 8(0.0075kg) F v (300m/s) 45N t 0.40s Δ = = = Δ 2 2 f 2 2 2 2 (46.1rad/s) (33.3rad/s) 236rad 2(2.15rad/s ) ω - ω = αΔθ-Δθ = = ( ) 2 2 2 2 t cp t 2 2 2 t 2 cp 2 a a a a r (4.65m)(0.745rad/s ) 3.46425m/s v (r ) a (4.65m) 1.25rad/s 7.26m/s r r a 8.05m/s = + = α = = ω = = = = = ! ! ! ! ! ! ! ! All of Ball #1’s potential energy is converted to translational (straight-line) kinetic energy, whereas Ball #2’s potential energy is shared between its translational and rotational kinetic energies. This means that the velocity of Ball #1’s center-of-mass will be greater and it will win the “race”!...
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This note was uploaded on 10/13/2011 for the course PHY 101 taught by Professor Ashkenkai during the Spring '08 term at FIU.

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Physics 101 Exam 2 - f,astronaut p p (75kg 25kg)v...

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