Question

When a particle of mass m is at (x,0), it is attracted toward the origin with a force whose magnitude is k/x^2 where k is some constant. If a particle starts from rest at x = b and no other forces act on it, calculate the work done on it by the time it reaches x = a, 0 < a < b.

Answer 1

Answer:

$W=k\frac{(b-a)}{ab}$

Explanation:

Given:

Force, $F={k}{x^2}$

Now we know,

work done, W = Force × Displacement

Now for the small displacement, dx the work done will be

W=F.dx

Now for the work done from 'a' to 'b' will be

$W=\int\limits^b_a {F} \, dx$

or

$W=\int\limits^b_a {\frac{k}{x^2}} \, dx$

now for the given condition 0<a<b , and the particle is moving from point b to the point 'a'

thus, the work done will be negative

therefore,

$W=\int\limits^a_b {\frac{k}{x^2}} \, (-dx)$

or

$W=-k\int\limits^a_b {\frac{1}{x^2}} \, dx$

or

$W=-k[\frac{1}{x}]^b_a$

or

$W=-k[\frac{1}{b}-\frac{1}{a}]$

or

$W=-k\frac{(a-b)}{ab}$

or

$W=k\frac{(b-a)}{ab}$  (Answer)