Homework Help: Physics

Posted by Cede on Wednesday, February 22, 2012 at 6:37pm.

I cant figure this problem out and it's asking for the value and units for the force. Please help me!

A 5.25 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t +(0.61(m/s^3)(t^3).

-What is the magnitude of the force F when 3.60 s?

Physics - Damon, Wednesday, February 22, 2012 at 8:30pm
F = m a
what is a?
d^2y/dt^2 = a
dy/dt = 2.80 +(3*.61)t^2
d^2y/dt^2 = a = (6*.61) t
F = m a = (5.25*6*.61)t
so when t = 3.6
F = 5.25 * 6 * .61 * 3.6 Newtons

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