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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...Asked by Jae
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?
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Answered by
Anonymous
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)
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)