{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

final 06 - Version 122/ABDCC nal 06 Turner(59070 This...

This preview shows pages 1–4. Sign up to view the full content.

Version 122/ABDCC – final 06 – Turner – (59070) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It takes 6 J of work to stretch a Hooke’s-law spring 6 . 95 cm from its unstressed length. How much the extra work is required to stretch it an additional 15 . 7 cm? 1. 18.7066 2. 27.7603 3. 8.36233 4. 6.84479 5. 57.7262 6. 16.2043 7. 4.97423 8. 12.552 9. 18.4614 10. 17.1564 Correct answer: 57 . 7262 J. Explanation: We begin by determining the value of k from the energy equation solved for k , k = 2 U x 2 = 2(6 J) (0 . 0695 m) 2 = 2484 . 34 J / m 2 We can now determine the energy at the ad- ditional displacement, U add = 1 2 k x 2 add = 1 2 2484 . 34 J / m 2 (0 . 0695 m + 0 . 157 m) 2 = 63 . 7262 J The extra work required is just the difference in energy between the two displacements. W = Δ U = U add U = 63 . 7262 J 6 J = 57 . 7262 J 002 10.0 points A 130 kg block is released at height h = 3 . 7 m as shown. The track is frictionless except for a portion of length 6 . 9 m. The block travels down the track, hits a spring of force constant k = 1537 N / m, and compresses it 2 . 2 m from its equilibrium position before coming to rest momentarily. The acceleration of gravity is 9 . 8 m / s 2 . μ 3 . 7 m 2 . 2 m 6 . 9 m 130 kg 1537 N / m Determine the coefficient of kinetic friction between surface and block over the 6 . 9 m track length. 1. 0.113105 2. 0.527975 3. 0.390973 4. 0.447849 5. 0.202683 6. 0.513716 7. 0.464463 8. 0.15323 9. 0.174255 10. 0.148437 Correct answer: 0 . 113105. Explanation: Let : m = 130 kg , k = 1537 N / m , h = 3 . 7 m , = 6 . 9 m , x = 2 . 2 m , and g = 9 . 8 m / s 2 . From conservation of energy Δ E = W f or k x 2 2 m g h = μ m g ℓ . Hence, μ = m g h k x 2 2 m g ℓ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 122/ABDCC – final 06 – Turner – (59070) 2 = (3 . 7 m) (1537 N / m) (2 . 2 m) 2 2 (130 kg)(9 . 8 m / s 2 ) (6 . 9 m) = 0 . 113105 . 003 10.0 points An 89 g autographed baseball rolls off of a 1 . 5 m high table and strikes the floor a hori- zontal distance of 1 m away from the table. 1 . 5 m 1 m How fast was it rolling on the table before it fell off? The acceleration of gravity is 9 . 81 m / s 2 . 1. 1.80831 2. 2.32282 3. 1.96384 4. 1.45683 5. 1.67806 6. 1.26259 7. 1.51614 8. 1.47816 9. 2.05896 10. 1.57338 Correct answer: 1 . 80831 m / s. Explanation: Let : Δ x = 1 m , Δ y = 1 . 5 m , and g = 9 . 81 m / s 2 . v x Δ y Δ x From the vertical motion, since v i,y = 0, Δ y = v i,y Δ t + 1 2 g t ) 2 = 1 2 g t ) 2 Δ t = radicalBigg 2 Δ y g From the horizontal motion, since a x = 0, Δ x = v i,x Δ t + 1 2 a x t ) 2 = v x Δ t v x = Δ x Δ t = radicalbigg g y Δ x = radicalBigg 9 . 81 m / s 2 2( 1 . 5 m) (1 m) = 1 . 80831 m / s . 004 10.0 points A uniform rod of mass 1 . 5 kg is 3 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 3 m from the center of mass of the rod. The rod is released from rest at the horizontal position.
Version 122/ABDCC – final 06 – Turner – (59070) 3 3 m 3 m 1 . 5 kg O 41 What is the angular acceleration of the rod at the instant the rod makes an angle of 41 with the horizontal? The acceleration of grav- ity is 9 . 8 m / s 2 , and the moment of inertia of the rod about its center of mass is 1 12 m ℓ 2 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 19

final 06 - Version 122/ABDCC nal 06 Turner(59070 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online