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final - Husain Zeena Final 1 Due 5:00 pm Inst H L Berk This...

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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. 52 kg μ = 0 . 4 F 48 m 20 m
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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 sin θ N = m g cos 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 the ramp, we have F = F g k + F k = m g [sin θ + μ k cos θ ] = (52 kg) (10 m / s 2 ) 20 52 + (0 . 4) 48 52 = 392 N .
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