How large a force parallel to a 30° incline is needed to give a 5.0 kg an acceleration 0.20 m/s^2 up the incline a) if friction is negligible? b) if the coefficient is 0.30?

Answer 1

weigh force normal to slope = m g cos 30
so friction force down slope = 0.30 m g cos 30

weight force down slope = m g sin 30

a ) F = m g sin 30
b ) F - m g sin 30 - 0.30 m g cos 30 = m a
so
F = 5 ( 0.5 g + 0.26 g + 0.20)