In the figure masses `400 kg` and `100 kg` are fixed.

(a) How much work must be done to move a `1 kg` mass from point `A` to point `B`?
(b) What is the minimum kinetic energy with which the `1 kg` mass must be projected from `A` to the right to reach the point `B`?

Answer

A, B

Explanation

`W = U_(f) - U_(i) = U_(B) - U_(A)`
`= mV_(B) - mV_(A)` (`V =` gravitational potential)
`= m(V_(B) - V_(A))`
`= (1) [-(100)/(2) - (400)/(8) + (100)/(8) + (400)/(2)] (6.67 xx 10^(-11))`
`= 7.5 xx 10^(-9) J`
(b) First let us find the point (say) `c` between `A` and `B` where field strength due to `400 kg` and `100 kg` masses is zero. Let its distance from `400 kg` is `r`.
Then, `(G(400))/(r^(2)) = (G(100))/((10 - r)^(2))`
`r = (20)/(3) m` and `(10 - r) = (10)/(3) m`
So, we have to move the body only from `A` to `C`. After that `100 kg` mass will pull `1 kg` mass by its own.
`:.` Minimum kinetic energy `= Delta U = U_(C) - U_(A)`
`= m(V_(C) - V_(A))`
`= (1) [-(400)/(20//3) - (100)/(10//3) + (100)/(8) + (400)/(2)] (6.67 xx 10^(-11))`
`= 8.17 xx 10^(-9)J`