Question

+ A machine carries a 7.0 kg package from an initial position of d ; = (0.8 m) î + (0.77 m)ị + (0.26 m)k at t = 0 to a final position of de = (7.5 m) î + (15.0m)ị + (10.2 m)k at t = 13.0 s. The constant force applied by the machine on the package is } = (2.0N) î + (7.0N)ị + (7.0N). For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package. f

Answer 1

Solution:

From the information given,

The displacement of the package,

d = d_f - d_i = [7.5 - 0.8]i + [15-0.77]j + [10.2-0.26]k

= [6.7 i + 14.23 j + 9.94 k]m

a) The work done on the package is dot product of d and f ,

W_net = d ^. F

= [6.7 i + 14.23 j + 9.94 k]m ^.[2.0 i + 7.0 j + 7.0 k ] N

= 182.59 J.

= 183 J.

b) The average power of the machines process,

P_avg = W_net/ $\Delta$ t

= 183/13

= 14.05 Watt.

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