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A machine carries a 3.0 kg package from an initial position of i = (0.50 m) + (0.75 m) + (0.20 m) at t = 0 to a final position of f = (7.80 m) + (12.5 m) + (7.40 m) at t = 11 s. The constant force applied by the machine on the package is = (2.00 N) + (4.00 N) + (6.00 N).

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Answer:

Answer:

mass of package, m = 3 kg

Explanation:

initial position, [tex]\overrightarrow{r_{1}}=0.5\widehat{i}+0.75\widehat{j}+0.2\widehat{k}[/tex]

Final position,

[tex]\overrightarrow{r_{2}}=7.8\widehat{i}+12.5\widehat{j}+7.4\widehat{k}[/tex]

time, t = 11 s

Force,

[tex]\overrightarrow{F}=2\widehat{i}+4\widehat{j}+6\widehat{k}[/tex]

Work done,

[tex]W = \overrightarrow{F}.\overrightarrow{d}[/tex]

where, d = r2 - r1

[tex]\overrightarrow{d}=\left ( 7.8-0.5 \right )\widehat{i}+\left ( 12.5-0.75 \right )\widehat{j}+\left ( 7.4-0.2 \right )\widehat{k}[/tex]

[tex]\overrightarrow{d}=7.3\widehat{i}+11.75\widehat{j}+7.2\widehat{k}[/tex]

W = [tex]\left (  2\widehat{i}+4\widehat{j}+6\widehat{k}\right ).\left (\overrightarrow{d}=7.3\widehat{i}+11.75\widehat{j}+7.2\widehat{k}  \right )[/tex]

W = 104.8 J

Power = Work / time

P = 104.8 / 11 = 9.53 Watt

Explanation:

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