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

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

Toppr Admin · Accepted Answer

Fg-m-x2192-Eg-1-10-i-x2212-j-10-i-x2212-j-N-10-i-x2212-10-jN-x2192-S-2-i-x2212-2-j-x2212-i-x2212-j-i-x2212-3-jmWork done-xA0-x2192-Fg-x2192-S-x2261-10-i-x2212-10-j-i-x2212-3-j-10-1-x2212-10-x2212-3-40J

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
$- 10 J$
$- 40 J$
$10 J$
$+ 40 J$

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

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−10 J−40 J10 J+40 J

Fg=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
$- 10 J$
$- 40 J$
$10 J$
$+ 40 J$

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