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A machine carries a 4.0 kg package from an initial position of di = (0.50 m) + (0.75 m) + (0.20 m) at t = 0 to a final position of df = (7.20 m) + (13.0 m) + (7.30 m) at t = 11 s. The constant force applied by the machine on the package is F = (2.00 N) + (4.00 N) + (6.00 N) . (a) For that displacement, find the work done on the package by the machine's force. _________ J (b) Also, find the power of the machine's force on the package. ________ W
A machine carries a 4.0 kg package from an initial position of di = (0.50 m) + (0.75 m) + (0.20 m) at t = 0 to a final position of df = (7.20 m) + (13.0 m) + (7.30 m) at t = 11 s. The constant force applied by the machine on the package is F = (2.00 N) + (4.00 N) + (6.00 N) . (a) For that displacement, find the work done on the package by the machine's force. _________ J (b) Also, find the power of the machine's force on the package. ________ W
Final answer:The work done by the machine's force on the package is 105 Joules and the power of the machine's force on the package is 9.55 Watts.Explanation:To answer your question, we first need to understand the concept of work and power in Physics.Workis defined as the dot product of the force and the displacement, i.e., W = F.d = (Fx*Dx + Fy*Dy + Fz*Dz)Power, on the other hand, is the rate of doing work or the amount of work per unit time.(a) The displacement of the package is (7.20 m - 0.50 m) i + (13.0 m - 0.75 m) j + (7.30 m - 0.20 m) k = (6.70 m) i + (12.25 m) j + (7.10 m) k. Therefore, the work done by the force on the package = F.d = (2.00 N)(6.70 m) + (4.00 N)(12.25 m) + (6.00 N)(7.10 m) = 13.4 J + 49.0 J + 42.6 J = 105 J.(b) Power = Work / Time = 105 J / 11 s = 9.55 W. So, the power of the machine's force on the package is 9.55 W.Learn more about Physics of Work and Power here:brainly.com/question/33726554#SPJ3...