A machine carries a 7.0 kg package from an initial position of d Overscript righ

Question

A machine carries a 7.0 kg package from an initial position of d Overscript right-arrow EndScripts Subscript i Baseline equals left-parenthesis 0.9 m right-parenthesis i Overscript EndScripts plus left-parenthesis 0.78 m right-parenthesis j Overscript EndScripts plus left-parenthesis 0.29 m right-parenthesis k Overscript EndScripts at t = 0 to a final position of d Overscript right-arrow EndScripts Subscript f Baseline equals left-parenthesis 9.5 m right-parenthesis i Overscript EndScripts plus left-parenthesis 12.0 m right-parenthesis j Overscript EndScripts plus left-parenthesis 8.2 m right-parenthesis k Overscript EndScripts at t = 16.0 s. The constant force applied by the machine on the package is Upper F Overscript right-arrow EndScripts equals left-parenthesis 6.0 N right-parenthesis i Overscript EndScripts plus left-parenthesis 6.0 N right-parenthesis j Overscript EndScripts plus left-parenthesis 6.0 N right-parenthesis k Overscript EndScripts. 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.

Explanation / Answer

Given that,

m = 7 kg

di = (0.9 i + 0.78 j + 0.29 k) m

df = (9.5 i + 12 j + 8.2 k) m

t = 16 s

F = (6 i + 6 j + 6 k) N

(a)

Wok done, W = F . (df - di)

W =  (6 i + 6 j + 6 k) . [ (9.5 i + 12 j + 8.2 k) -  (0.9 i + 0.78 j + 0.29 k) ]

W =  (6 i + 6 j + 6 k) . (8.6 i + 11.22 j + 7.91 k)

W = 166.38 J

(b)

average power of the machine's force on the package,

P = W / t

P = 166.38 / 16

P = 10.39 W

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