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Standard X

Physics

Potential Energy of an Object at a Height

The gravitati...

Question

The gravitational intensity a region is $10 (^i -^j) N / k g$ . The work done by the gravitational force to shift slowly a particle of mass $1 K g$ from point $(1 m, 1 m)$ to a point $(2 m, - 2 m)$ is

10 J

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- 10 J

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- 40 J

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+ 40 J

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Solution

The correct option is D $+ 40 J$
$F_{g} = m_{g} = 1 (10 (^i -^j)) = 10 (^i -^j) N = 10^i - 10 {^j}_{N}$
$\to S = (2^i - 2^j) - (^i -^j) =^i - 3^j m$
Work done, $_{g} . \to S \equiv (10^i - 10^j) . (^i - 3^j) = 10 (1) + (- 10) (- 3) = + 40 J$

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The gravitational intensity 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

The correct option is D +40 JFg=m→Eg=1(10(^i−^j))=10(^i−^j)N=10^i−10^jN→S=(2^i−2^j)−(^i−^j)=^i−3^jmWork done, →Fg.→S≡(10^i−10^j).(^i−3^j)=10(1)+(−10)(−3)=+40J

The gravitational intensity a region is $10 (^i -^j) N / k g$ . The work done by the gravitational force to shift slowly a particle of mass $1 K g$ from point $(1 m, 1 m)$ to a point $(2 m, - 2 m)$ is

The correct option is D $+ 40 J$
$F_{g} = m_{g} = 1 (10 (^i -^j)) = 10 (^i -^j) N = 10^i - 10 {^j}_{N}$
$\to S = (2^i - 2^j) - (^i -^j) =^i - 3^j m$
Work done, $_{g} . \to S \equiv (10^i - 10^j) . (^i - 3^j) = 10 (1) + (- 10) (- 3) = + 40 J$