Let the acceleration of the system be a and the tension in the string be T.

For particle P on the table, applying Newton's second law:

\(T = 1.7a\)

For particle Q hanging vertically, applying Newton's second law:

\(0.3g - T = 0.3a\)

Substitute the expression for T from the first equation into the second equation:

\(0.3g - 1.7a = 0.3a\)

Combine and solve for a:

\(0.3g = 2a\)

\(a = \frac{0.3g}{2}\)

Using \(g = 9.8 \text{ m/s}^2\):

\(a = \frac{0.3 \times 9.8}{2} = 1.47 \text{ m/s}^2\)

Rounding to one decimal place, \(a = 1.5 \text{ m/s}^2\).

Substitute \(a = 1.5 \text{ m/s}^2\) back into the equation for T:

\(T = 1.7 \times 1.5 = 2.55 \text{ N}\)