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HW6 - homework 06 – GAUTHIER JOSEPH – Due 1:00 am 1...

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Unformatted text preview: homework 06 – GAUTHIER, JOSEPH – Due: Oct 12 2007, 1:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points The figure below shows a rough semicircu- lar track whose ends are at a vertical height h . A block placed at point P at one end of the track is released from rest and slides past the bottom of the track. h P What height on the far side of the track does the block reach if the coefficient of kinetic friction is 0 . 4 . 1. h ′ = 0 . 6 h 2. h ′ = 0 . 4 h 3. It is between zero and h ; the exact height depends on how much energy is lost to fric- tion. correct 4. h ′ = 0 . 6 h 2 5. h ′ = 0 . 6 h 2 π 6. h ′ = h 7. h ′ = 0 . 4 h 2 π 8. h ′ = 0 . 4 h 2 Explanation: Because the surface of the track is rough, the block will lose some energy to friction ( W < 0). Use the work energy theorem E f − E i = W, we know that the height h ′ to which the block rises on the other side of the track should be E f − E i = W mg h ′ − mg h = W < mg h ′ < mg h h ′ < h Therefore, < h ′ < h . Question 2, chap 8, sect 1. part 1 of 6 10 points A heavy ball swings on a string in a circular arc of radius 1 m. Q 1 m Q ′ P 2 7 ◦ The two highest points of the ball’s trajec- tory are Q and Q ′ ; at these points the string is ± 27 ◦ from the vertical. Point P is the lowest point of the ball’s trajectory where the string hangs vertically down. The acceleration of gravity is 9 . 8 m / s 2 . What is the ball’s speed at the point P ? Ne- glect air resistance and other frictional forces. Correct answer: 1 . 4616 m / s (tolerance ± 1 %). Explanation: Let : g = 9 . 8 m / s 2 , m = 4 kg , ℓ = 1 m , and θ = 27 ◦ . The diagram for the velocity and accelera- tion of the ball at point P is as follows, x y v a homework 06 – GAUTHIER, JOSEPH – Due: Oct 12 2007, 1:00 am 2 The difference of gravitational potential en- ergy of the ball between point Q and point P is Δ U = U Q − U P = mg ℓ (1 − cos θ ). The velocity of the ball at point Q is zero, so K P + U P = K Q + U Q K P + U P = 0 + U Q . Therefore 1 2 mv 2 P = K P = U Q − U P = mg ℓ (1 − cos θ ) v P = radicalbig 2 g ℓ (1 − cos θ ) = radicalBig 2 (9 . 8 m / s 2 ) (1 m) (1 − cos 27 ◦ ) = 1 . 4616 m / s . Question 3, chap 8, sect 1. part 2 of 6 10 points What is the magnitude of the ball’s accel- eration at the point P ? Correct answer: 2 . 13627 m / s 2 (tolerance ± 1 %). Explanation: At point P , the centripetal acceleration is the total acceleration: a P = v 2 P ℓ = (1 . 4616 m / s) 2 1 m = 2 . 13627 m / s 2 . Question 4, chap 8, sect 1. part 3 of 6 10 points What is the ball’s speed at the point Q ? 1. bardbl v Q bardbl = bardbl v p bardbl 2. bardbl v Q bardbl < 0 m/s 3. bardbl v Q bardbl = 0 m/s correct 4. 0 m / s < bardbl v Q bardbl < bardbl v p bardbl 5. bardbl v Q bardbl > bardbl v p bardbl Explanation: Since Q is one of the highest points, the velocity at Q is zero ....
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HW6 - homework 06 – GAUTHIER JOSEPH – Due 1:00 am 1...

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