Question

If a force $F= -\frac{K}{X^2}(x \neq 0)$ acts as a particle along the x-axis. Find the work done by the force in displacing the particle from x = +a to x = +2a. Take K as a positive constant.

Answer 1

workdone is the dot product of force and displacement. e.g., $W=\vec{F}.\vec{s}$

when force is function of position e.g.,y=F(x)
then, workdone , W = F(x).dx

so, W = $\int\limits^{2a}_a{F(x)}\,dx$

= $\int\limits^{2a}_a{-\frac{K}{x^2}}\,dx$

= $-K\left[\frac{1}{-x}\right]^{2a}_a$

= $-K\left[\frac{1}{2a}-\frac{1}{a}\right]$

= $\frac{K}{2a}$

hence ,work done by the force in displacing the particle from x = +a to x = +2a is $\frac{K}{2a}$

Answer 2

Hii dear,

● Answer-
W = k/2a

● Explaination-
Work done is given by
Work = Force × displacement
W = F.s

Energy stored at instant x = a,
E1 = -k/x^2 × x
E1 = -k/x
E1 = -k/a

Energy stored at instant x = 2a,
E1 = -k/x^2 × x
E1 = -k/x
E1 = -k/2a

Total work done,
W = E2 - E1
W = -k/2a + k/a
W = k/2a

Thus total work done is k/2a...

Hope this is helpful...