final - Husain, Zeena – Final 1 – Due: Dec 11 2003,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Husain, Zeena – Final 1 – Due: Dec 11 2003, 5:00 pm – Inst: H L Berk 1 This print-out should have 25 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points The square of the speed of an object undergo- ing a uniform acceleration a is some function of a and the displacement s , according to the expression given by v 2 = ka m s n , where k is a dimensionless constant. Using dimensional analysis 1. v 2 = k s 3 a 2. v 2 = k a 2 s 3. v 2 = k a 2 s 4. v 2 = k a s 5. v 2 = k s a 6. v 2 = k a 3 s 7. v 2 = k a s correct 8. v 2 = k √ a s 9. v 2 = k a 2 s 2 10. v 2 = k r a s Explanation: Solution: If the dimensions (length, mass, time) of the right hand side of a formula do not equal those of the left hand side, you know right off that formula is NOT correct. The dimensions of velocity are [ v ] = L T = L T- 1 . So the dimensions of the left hand side of the formula in question are [ v 2 ] = L 2 T 2 = L 2 T- 2 . On the right hand side of the formula, there is k , which is dimensionless, multiplied by the acceleration [ a ] = L T 2 = L T- 2 and the displacement [ s ] = L . What exponent will give the dimensions of v 2 ? [ v 2 ] = [ k a m s n ] . L 2 T- 2 = ( L T- 2 ) m L n = L m + n T- 2 m . For the dimensions of the two sides of the formula to be the same, the exponent of L has to be equal to 2 and the exponent of T has to be- 2 m + n = 2 (1)- 2 m =- 2 . (2) Equation 2 implies that m = 1 . Substitute this into Eq. 1 to get 1 + n = 2 n = 1 . Therefore v 2 = k a s . 002 (part 1 of 1) 10 points Given: Pretend you are on a planet similar to Earth where the acceleration of gravity is approximately 10 m / s 2 . The coefficient of kinetic friction between the plane and the block is 0 . 4. A block of mass 52 kg lies on an inclined plane, as shown below. The horizontal and vertical supports for the plane have lengths of 48 m and 20 m, respectively. 5 2 k g μ = . 4 F 48 m 20m Husain, Zeena – Final 1 – Due: Dec 11 2003, 5:00 pm – Inst: H L Berk 2 The magnitude of the force ~ F necessary to pull the block up the plane with constant speed is most nearly 1. k ~ F k ≈ 250 N 2. k ~ F k ≈ 270 N 3. k ~ F k ≈ 114 N 4. k ~ F k ≈ 110 N 5. k ~ F k ≈ 392 N correct 6. k ~ F k ≈ 344 N 7. k ~ F k ≈ 330 N 8. k ~ F k ≈ 100 N 9. k ~ F k ≈ 190 N 10. k ~ F k ≈ 290 N Explanation: Given : x = 48 m , y = 20 m , s = p x 2 + y 2 = q (48 m) 2 + (20 m) 2 = 52 m , m = 52 kg , g ≈ 10 m / s 2 , a = 0 m / s 2 , θ = arctan ‡ y x · = 22 . 6199 ◦ sin θ = y s = 20 52 cos θ = x s = 48 52 . s θ x y Consider the free body diagram for the block m g s i n θ N = m g c o s θ μ N F m g Basic Concepts: Parallel to the ramp F net = F g k + F k- F = m a = 0 , F g k = m g sin θ , F k = μ k N , so F = F g k + F k . (1) Perpendicular to the ramp F net = N- F g ⊥ = 0 , so N = m g cos θ . (2) Solution: Using Eqs. 1 and 2, parallel to theUsing Eqs....
View Full Document

This test prep was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 11

final - Husain, Zeena – Final 1 – Due: Dec 11 2003,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online