Question

A particle in the xy plane travels along a spiral path C beginning at a point P that is 8 units from the origin and ending at a point Q that is 2 units from origin.

The particle makes 2.5 revolutions aroung the origin along the way.

What is work done by the gravitational field

F(x, y) = i + $\frac{-y}{(x^2 + y^2)^(3/2)}$ j in moving the particle along its path?

(x2 + y2) 3/2)

Answer 1

First , the gravitational field that is given is conservstive in nature .

If force is conservstive in nature , then work done depends on only initial & final position and also does not depends on path taken by the particle.

Here in this question path taken by particle is spiral.

But we can also calculated the work done if particle comes along X axis from point (8,0) to (2,0) , as work due to conservative force does not depend on path taken

Now if we take particle is moving along X axis , then y = 0

If we put y= 0 in the field , then

F(x,0) = -1/x^2 i + 0

I.e only force along X axis will act

Now work done by this force is integration of Fx from x = 8 to x = 2

$Work = \int_{8}^{2} F(x)dx$​​​​​​

$Work = \int_{8}^{2} (-1/{x}^2)dx$

On solving , we get

Work = 3/8

Option A is answer