Answer
6.0 J
Work Step by Step
$W=\int^{x_{f}}_{x_{i}}F(x)dx=\int^{3.0}_{1.0}\frac{a}{x^{2}}dx$
$=a\int^{3.0}_{1.0}\frac{1}{x^{2}}dx=a\times[-\frac{1}{x}]^{3.0}_{1.0}= a(-\frac{1}{3.0\,m}-(-\frac{1}{1.0\,m}))$
$= 9.0\,N\cdot m^{2}\times\frac{2}{3.0\,m}= 6.0\,J$
