- The gravitational force acting down the slope: \(F_{g\parallel} = mgsin(\theta)\), where \(m = 3\,kg\), \(g = 9.8\,m/s^2\), and \(\theta = 30^\circ\).
- The normal force acting perpendicular to the slope: \(N = mgcos(\theta)\).
- The frictional force opposing
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