OLDHW15 - oldhomewk 15 – PAPAGEORGE, MATT – Due: Feb 21...

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Unformatted text preview: oldhomewk 15 – PAPAGEORGE, MATT – Due: Feb 21 2008, 4:00 am 1 Question 1, chap 8, sect 5. part 1 of 1 10 points A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ . At what rate does the child do work on the box? 1. P = μmg v 2. P = μmg v correct 3. P = v μmg 4. P = mg v 5. P = μmv 2 Explanation: Since the box moves horizontally with con- stant speed, the force exerted by the child is equal to the frictional force on the box μmg . Thus, the power generated by the child is P = F v = μmg v . Question 2, chap 6, sect 1. part 1 of 1 10 points A box of mass m with an initial velocity of v slides down a plane, inclined at θ with respect to the horizontal. The coefficient of kinetic friction is μ . The box stops after sliding a distance x . m μ k v θ How far does the box slide? 1. x = v 2 2 g ( μ cos θ + sin θ ) 2. x = v 2 g ( μ sin θ- 2 cos θ ) 3. x = v 2 2 g sin θ 4. x = v 2 g ( μ sin θ + cos θ ) 5. x = v 2 2 g ( μ cos θ- sin θ ) correct 6. x = v 2 g (sin θ- μ cos θ ) 7. x = v 2 2 g μ cos θ 8. x = v 2 2 g ( μ sin θ- cos θ ) 9. x = v 2 2 g (sin θ- μ cos θ ) 10. x = v 2 2 g ( μ sin θ + cos θ ) Explanation: Basic Concepts: Motion under constant force W = vector F · vectors P = Δ W Δ t P = dW dt = vector F · vectorv . Solution: The net force on the block parallel to the incline is F net = F mg sin θ- F f , where F f is the friction force. Thus, Newton’s equation for the block reads ma = mg sin θ- F f = mg sin θ- μ N = mg (sin θ- μ cos θ ) a = g (sin θ- μ cos θ ) , where N = mg cos θ . To find the distance the block slides down the incline, use v 2 = v 2 + 2 a ( x- x ) , oldhomewk 15 – PAPAGEORGE, MATT – Due: Feb 21 2008, 4:00 am 2 valid for a body moving with a constant accel- eration. Since x = 0 and v f = 0 (the block stops), we get x =- v 2 2 a =- v 2 2 g (sin θ- μ cos θ ) = v 2 2 g ( μ cos θ- sin θ ) . Question 3, chap 8, sect 5. part 1 of 1 10 points Water flows over a section of Niagara Falls at a rate of 1 . 9 × 10 6 kg / s and falls 67 m. The acceleration of gravity is 9 . 8 m / s 2 . What is the power wasted by the waterfall? Correct answer: 1 . 24754 × 10 9 W (tolerance ± 1 %). Explanation: P = W t = mg h t = ρg h = (1 . 9 × 10 6 kg / s) (9 . 8 m / s 2 ) (67 m) = 1 . 24754 × 10 9 W Question 4, chap 8, sect 5. part 1 of 4 10 points A particle of mass m moves along the x axis. Its position varies with time according to x = (5 m / s 3 ) t 3- (7 m / s 2 ) t 2 . What is the velocity of the particle at any time t ? 1. v = (27 m / s 2 ) t 2- (14 m / s) t 2. v = (15 m / s 2 ) t 2- (14 m / s) t correct 3. v = (24 m / s 2 ) t 2- (16 m / s) t 4. v = (27 m / s 2 ) t 2- (4 m / s) t 5. v = (6 m / s 2 ) t 2- (12 m / s) t 6. v = (24 m / s 2 ) t 2- (8 m / s) t 7. v = (12 m / s 2 ) t 2- (16 m / s) t 8. v = (18 m / s 2 ) t 2- (12 m / s) t 9. v = (6 m / s 2 ) t 2- (6 m / s)...
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.

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OLDHW15 - oldhomewk 15 – PAPAGEORGE, MATT – Due: Feb 21...

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