Question

A spaceship comes upon a 92kg solid gold space rock and wishes to bring it into its storage bay. It has a tractor beam that exerts a force 902/x^4 N towards the spaceship, where x is the distance from the spaceship. So the farther away the asteroid is, the weaker the pull. The tractor beam is effective from 7m to 2m. If the object is beyond this range, there is no force, and the tractor beam ceases to function at shorter distances. If the rock was at rest relative to the spaceship at the start of the problem, what is its speed when it enters the spaceship?

Answer 1

The speed of the rock is 4.38 m/s.

Solution:

As per the question:

Mass of the gold space rock, m = 92 kg

Force exerted by the tractor beam,
`F_(B) = (902)/(x^(4))`

Now,

Suppose that the spaceship started to apply force when it reaches a distance of 7 m until it gets close to 2m distance, in this case the energy, E can be given as:

`Energy, E = \int_(6)^(2)F.dx`

`E = 902* [- (1)/(x^(3))]_(6)^(2) = 885.29\ J`

Now,

To calculate the speed, the gain in Kinetic energy of the rock is given by:

`|E| = (1)/(2)mv^(2)`

`v = \sqrt{(2E)/(m)}`

`v = \sqrt{(2* 885.29)/(92)} = 4.38\ m/s`