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April 19, 2014

April 19, 2014

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

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 8:30pmF = 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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