Question

PRINTER VERSION BACK NEXT Chapter 07, Problem 041 SN incorrect. with x in meters A single force acts on a particle-like object of mass m kg in such a way that the position of the object as a function of time is given by x 3.4 and t in seconds. Find the work done on the object by the force between 0 and time t. Express your answer in terms of the variables given. m(61-8)(31-(41) +) Edit Click if you would like to show Work for this question: Open Show Work

Answer 1

59: Given d = 3t- 4+² + t² work done = Ex di = mxaxd a=dv-d caw at de 13t-4tt?) - a (3-8t+3+²) - - 8 +66 = (6-8) Now W- mx (64-8) ( 3t-4t² t). your answer is almost correct except that you have place square for (4t), "But. square is only for t. itis 4th Not (4t) ².

Given F = 16 NT + 8 6 Nit 8 INK (F) = √ (1.0² + (8.6)²+(8-17² = 142.13 11.9218 N d; = a.li +8.0 ] +0k, di= 9.6 T + 6:40 ] + 7.8m E d = d-di = (9.1 - 966) T +(8:0–64) I+ (1.0 -7.8) F = -0.5i + 1.65 – 6.85 1d) -(0.5) + (116)2 +16-3)? = √ 49.05 = 7:0035m work done - Fod = 11:9218x7.0035 = 183-495 J 6 power = workdons time = 33.495 1.6 - 52.184 Watts

•Coso = di di Idilldil = ( 911 +81 +1E) • (9.61 + 6.4 1 + 7185) 9.3 + 8 + 1² x 916 + 6y + 782 9.1x9.6 + 8x boy et 1x708 V 14781 & 193.96. - 87.36 + 51.2 + 708 12.15 77X 13.926 coso 00864636 o = cos (0.864636) - 30.158°
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