A body moves from an initial point (A) (2m, 3m, 4m) to a final point (B) (4m, 2m, 1m) in 9 seconds under the action of forcesF¯1=3i^+4j^+6k^N,F¯2=2i^-3k^N,F¯3=4i^-4j^NAverage power developed in this time is ___

F¯net =F¯1+F2+F¯3=9i^+0j^+3kN^b) r1=2i^+3j^+4k^mr1=2i^+3j^+4k^ m d¯=r2-r1=2i^-j^-3k^ c) W=F¯·d¯=9(2)+0(-1)+3(-3)=9 J d) Average power=Total workTotal time=9J9s=1 Watt

A body moves from an initial point (A) (2m, 3m, 4m) to a final point (B) (4m, 2m, 1m) in 9 seconds under the action of forcesF¯1=3i^+4j^+6k^N,F¯2=2i^-3k^N,F¯3=4i^-4j^NAverage power developed in this time is ___

F¯net =F¯1+F2+F¯3=9i^+0j^+3kN^b) r1=2i^+3j^+4k^mr1=2i^+3j^+4k^ m d¯=r2-r1=2i^-j^-3k^ c) W=F¯·d¯=9(2)+0(-1)+3(-3)=9 J d) Average power=Total workTotal time=9J9s=1 Watt

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A body moves from an initial point (A) (2m, 3m, 4m) to a final point (B) (4m, 2m, 1m) in 9 seconds under the action of forces
${\bar{F}}_{1} = 3 \hat{i} + 4 \hat{j} + 6 \hat{k} N, {\bar{F}}_{2} = 2 \hat{i} - 3 \hat{k} N, {\bar{F}}_{3} = 4 \hat{i} - 4 \hat{j} N$
Average power developed in this time is ___

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0.5 W

1 W

1.5 W

2 W

answer is B.

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Detailed Solution

${\bar{F}}_{net} = {\bar{F}}_{1} + F_{2} + {\bar{F}}_{3} = 9 \hat{i} + 0 \hat{j} + 3 \hat{k N}$

b) $r_{1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} m$
$r_{1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} m$
$\bar{d} = r_{2} - r_{1} = 2 \hat{i} - \hat{j} - 3 \hat{k}$
c) $W = \bar{F} \cdot \bar{d} = 9 (2) + 0 (- 1) + 3 (- 3) = 9 J$
d) $Average power = \frac{Total work}{Total time} = \frac{9J}{9s} = 1 Watt$