posted by Cede .
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?
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