To find the tension in the string, we resolve the forces parallel to the line of greatest slope. The weight component along the slope is given by \(W \sin \alpha\), where \(W = 5.1\) N and \(\sin \alpha = \frac{8}{17}\).

The tension \(T\) in the string has a component \(T \cos \beta\) along the slope, where \(\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \frac{24}{25}\).

Equating the components along the slope, we have:

\(T \cos \beta = W \sin \alpha\)

\(T \left(\frac{24}{25}\right) = 5.1 \left(\frac{8}{17}\right)\)

\(T = \frac{5.1 \times 8 \times 25}{17 \times 24}\)

\(T = 2.5 \text{ N}\)