The magnitude of the force exerted on the 3.0-kg block by the 2.0-kg block is 18 N.
The given parameters;
- applied force, F = 30 N
- inclination of the plane, θ = 30⁰
The net force on the two blocks is calculated by applying Newton's second law of motion as shown below;
[tex]\Sigma F_x = ma \\\\F + m_2gcos \theta \ - F_3_n= F_{net}\\\\ F + m_2gcos \theta - m_3g = F_{net}\\\\(30) + (2\times 9.8 \times cos\ 30) - (3\times 9.8) = F_{net}\\\\17.6 \ N = F_{net}\\\\F_{net} \approx 18 \ N[/tex]
Thus, the magnitude of the force exerted on the 3.0-kg block by the 2.0-kg block is 18 N.
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