The gravitational intensity in a region is 10( hati- hatj)N/kg. The work done by the gravitational force to shift slowly a particle of mass 1 Kg from point (Im, Im) to a point (2m, -2m) is

Question

The gravitational intensity in a region is 10( hati- hatj)N/kg. The work done by the gravitational force to shift slowly a particle of mass 1 Kg from point (Im, Im) to a point (2m, -2m) is (A) 10J (B) - 10J (C) - 40J (D) + 40J

Tardigrade Admin · Accepted Answer

Fg = m Eg = 1(10 ( hati - hatj)) = 10 ( hati - hatj) N = 10 hati - 10 hatj barS = (2 hati - 2 hatj) - ( hati - hatj) = hati - 3 hatj m Work done = Fg .S = (10 hati - 10 hatj ).( hati - 3 hatj ) = 10(1) + (-10)(-3)

Q. The gravitational intensity in a region is $10 (\hat{i} - \hat{j}) N / k g .$ The work done by the gravitational force to shift slowly a particle of mass 1 Kg from point (Im, Im) to a point (2m, -2m) is

10J

- 10J

- 40J

+ 40J

$F_{g} = m E_{g} = 1 (10 (\hat{i} - \hat{j})) = 10 (\hat{i} - \hat{j}) N = 10 \hat{i} - 10 \hat{j}$
$\overset{ˉ}{S} = (2 \hat{i} - 2 \hat{j}) - (\hat{i} - \hat{j}) = \hat{i} - 3 \hat{j} m$
Work done = $F_{g} . S = (10 \hat{i} - 10 \hat{j}) . (\hat{i} - 3 \hat{j})$
= 10(1) + (-10)(-3)

Q. The gravitational intensity in 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 (Im, Im) to a point (2m, -2m) is

10J

- 10J

- 40J

+ 40J

Fg​=mEg​​=1(10(i^−j^​))=10(i^−j^​)N=10i^−10j^​ Sˉ=(2i^−2j^​)−(i^−j^​)=i^−3j^​m Work done = Fg​​.S=(10i^−10j^​).(i^−3j^​) = 10(1) + (-10)(-3)

Q. The gravitational intensity in a region is $10 (\hat{i} - \hat{j}) N / k g .$ The work done by the gravitational force to shift slowly a particle of mass 1 Kg from point (Im, Im) to a point (2m, -2m) is

$F_{g} = m E_{g} = 1 (10 (\hat{i} - \hat{j})) = 10 (\hat{i} - \hat{j}) N = 10 \hat{i} - 10 \hat{j}$
$\overset{ˉ}{S} = (2 \hat{i} - 2 \hat{j}) - (\hat{i} - \hat{j}) = \hat{i} - 3 \hat{j} m$
Work done = $F_{g} . S = (10 \hat{i} - 10 \hat{j}) . (\hat{i} - 3 \hat{j})$
= 10(1) + (-10)(-3)