Question

The gravitational intensity at a region is 10(i-j)N/Kg. The work done by the gravitational force to shift slowly a particle of mass 1 Kg from point (1m,1m) to a point (2m, -2m) is what?

Answer 1

Answer:

$Fg=mEg=1(10(i^−j^))=10(i^−j^)N=10i^−10j^N</p><p>S=(2i^−2j^)−(i^−j^)=i^−3j^m</p><p>Work done, Fg.S≡(10i^−10j^).(i^−3j^)=10(1)+(−10)(−3)=+40J</p><p></p><p>$

Explanation:

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