How large a force parallel to a 300 incline is needed to give a 5.0 kg an acceleration 0.20 m/s2 up the incline a) if friction is negligible? b) if the coefficient is 0.30?

Answer 1

I suspect you mean 30 degrees
sin 30 = 0.500
cos 30 = 0.866
g = 9.81 m/s^2
-----------------------------------------
Normal force on ramp = m g cos 30 = 5 * 9.81 * 0.866 = 42.5 Newtons
Component of weight down ramp = m g sin 30 = 24.5 Newtons
------------------------------------------------------
without friction:F up ramp - 24.5 = m a
F up ramp = 24.5 + 5 *0.20 = 25.5 Newtons
----------------------------------------------------
with friction
F up ramp = 24.5 + 5 * 0.20 + 0.3 * 42.5 = 25.5 + 12.75 = 38.3 Newtons