A force F=-k / x2(x ≠ 0) acts on a particle in x-direction. Find the work done by the force in displacing the particle from x=-a to x=2 a.

Question

A force F=-k / x2(x ≠ 0) acts on a particle in x-direction. Find the work done by the force in displacing the particle from x=-a to x=2 a. (A) 3 k / 2 a (B) 4 k / a2 (C) -3 k / 2 a2 (D) (-9 k/a2)

Tardigrade Admin · Accepted Answer

Work done by the force in displacing the particle from x=-a to x=2 a will be W=∫ F d x=∫ limitsx=-ax=2 a(-(k/x2)) d x =[(k/x)]-a2 a =(k/2 a)-(k/(-a))=(3 k/2 a)

Q. A force $F = - k / x^{2} (x \neq = 0)$ acts on a particle in $x$ -direction. Find the work done by the force in displacing the particle from $x = - a$ to $x = 2 a$ .

$3 k /2 a$

$4 k / a^{2}$

$- 3 k /2 a^{2}$

$\frac{- 9 k}{a ^{2}}$

Work done by the force in displacing the particle from $x = - a$ to $x = 2 a$ will be
$W = \int F d x = x = - a \int x = 2 a (- \frac{k}{x ^{2}}) d x$
$= [\frac{k}{x}]_{- a}^{2 a}$
$= \frac{k}{2 a} - \frac{k}{( - a )} = \frac{3 k}{2 a}$

Q. A force F=−k/x2(x=0) acts on a particle in x-direction. Find the work done by the force in displacing the particle from x=−a to x=2a.

3k/2a

4k/a2

−3k/2a2

a2−9k​