This force is the component of the gravitational force acting along the incline, which is \( F = m \cdot g \cdot \sin(\theta) \), where \( m = 2.0 \mathrm{~kg} \), \( g = 9.8 \mathrm{~m/s^2} \), and \( \theta = 20^{\circ} \). So, \( F = 2.0 \mathrm{~kg} \cdot 9.8
Show more…