exam1_2007s_sol

exam1_2007s_sol - Physics 221 2007S Exam 1 Solutions [1]If...

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Physics 221 2007S Exam 1 Solutions Page 1 of 21 [1]If a 200kg box is subject to a net force of magnitude 100N, what is the magnitude of its acceleration? (A) 0.25 2 m/s (B) 0.50 2 (C) 1.0 2 (D) 2.0 2 (E) 4.0 2 Answer (B): Using Newton’s second law 2 100N m 0.5 200kg s F a m == = [2] A car is traveling a circular track at a constant speed of 20m/s. If the magnitude of the acceleration of the car is 2 1 m/s , what is the radius of the track? (A) 50m (B) 100m (C) 200m (D) 400m (E) 800m Answer (D): Centripetal acceleration is given by 2 v a R = so 2 2 2 (20 ) 400 (1 ) m s m s v R m a = .
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Physics 221 2007S Exam 1 Solutions Page 2 of 21 [3] An elevator has a mass of 600kg, not including passengers. The elevator is designed to ascend, at a constant speed, a vertical distance of 20.0m (i.e. five floors) in 16.0s, and is driven by a motor that can provide up to 30kW of power. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has a mass of 65kg. (A) 28 (B) 24 (C) 20 (D) 16 (E) 12 Answer (A): The force that the motor must exert on the elevator is equal to the weight of the elevator since it is moving at a constant speed. Since PF v = the force exerted by the motor and hence the weight of the elevator is / (30000 ) /(20 /16 ) 24000 WFP v W m s N == = = The mass of the elevator is thus 2 m s / (24000N) /(9.8 ) 2449kg mWg = From the problem, the number of passengers n is ( 600kg) / 65kg (2449kg 600kg) / 65kg=28.4 nm =− = The limit is thus 28 passengers. [4] A person pulls a 20-kg block up a ramp of height h = 10 m and angle θ = 20 ° at constant speed. The rope makes an angle φ = 35 ° with the ramp as shown and has a tension of 160N . Find the work done by kinetic friction. (A) 661J (B) 740J (C) 980J (D) 1880J (E) 5800J Answer (D): The length of the ramp is o 10m 29.3 sin sin 20 h x m θ Δ= = = The work done by the tension is o cos (160N)(29.3m)cos(35 ) 3840J T WT x ϕ = = The work done by gravity is 2 m s (20kg)(9.8 )(10m) 1960J g Wm g h . The normal force does no work since it is perpendicular to the ramp. By the work energy theorem, the total work on the block is zero because its kinetic energy does not change as the block is moving at a constant speed. Thus 0 ( ) (( 3840J) ( 1960J)) 1880J net g T k kT g W WWW WW W == + + + =− + + −
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Physics 221 2007S Exam 1 Solutions Page 3 of 21 [5] A bullet of mass 1g is fired at a block of wood traveling at a speed of 300 m/s. It comes to rest 60 μ s after hitting the surface of the wood. If you assume that the wood slows down the bullet exerting a constant force against the bullet’s motion, what is the magnitude of that force? (A) 2500N (B) 5000N (C) 7500N (D) 10000N (E) 25000N Answer (B): The acceleration of the bullet is (change in velocity)/(time interval) so 6 2 (300 ) /(60 ) 5.0 10 mm as ss μ == × in the direction against the motion of the bullet. From Newton’s second law, the force on the bullet is 36 2 m (10 kg)(5.0 10 ) 5000N s Fm a × = [6]A lady bug and a gentleman bug are sitting on a circular turntable with radius 50cm which is rotating clockwise at a constant angular velocity. The ladybug is at the rim of the turntable while the gentleman bug is half way between the center and the rim. If the lady bug is moving at a speed of 0.10m/s, what is the magnitude of the angular velocity of the gentleman bug?
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This note was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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exam1_2007s_sol - Physics 221 2007S Exam 1 Solutions [1]If...

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