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Unformatted text preview: homework 19 BAUTISTA, ALDO Due: Mar 8 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. The graph below shows the force on an object of mass M as a function of time. Time (s) Force(N) 1 2 3 4 10 10 For the time interval 0 to 4 s, the total change in the momentum of the object is 1. p = 40 kg m / s . 2. p = 20 kg m / s . 3. Indeterminable unless the mass M of the object is known 4. p = 20 kg m / s . 5. p = 0 kg m / s . correct Explanation: The Newtons second law of motion, in one dimension, is F = M a = M dv dt . From this, we obtain F dt = M dv , = ( M v ) = Z F dt, where M v is the momen tum of the object. So from the graph above, the change in the momentum is zero . Question 2 Part 1 of 2. 10 points. A 0.47 kg object is at rest. A 2.74 N force to the right acts on the object during a time interval of 1.39 s. a) What is the velocity of the object at the end of this time interval? Correct answer: 8 . 1034 m / s (tolerance 1 %). Explanation: Let to the right be positive: Let : m = 0 . 47 kg , F 1 = 2 . 74 N , and t 1 = 1 . 39 s . Since v i = 0 m/s, ~ F t = m~v f m~v i = m~v f v f, 1 = F t 1 m = (2 . 74 N) (1 . 39 s) . 47 kg = 8 . 1034 m / s , which is directed to the right. Question 3 Part 2 of 2. 10 points. At the end of this interval, a constant force of 4.23 N to the left is applied for 3.26 s. b) What is the velocity at the end of the 3.26 s? Correct answer: 21 . 2366 m / s (tolerance 1 %). Explanation: Let : v i, 2 = 8 . 1034 m / s , F 2 = 4 . 23 N , and t 2 = 3 . 26 s ....
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 Spring '08
 Turner
 Force, Mass, Work

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