Question

A 5.50 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) =(2.80 m/s)t +(0.61 m/s^3)t^3.
Required:
a. What is the magnitude of the force F when 4.10s ?
b. Is the magnitude's unit N but the system doesn't accept it?

Answer 1

Answer:

Explanation:

position

y(t) = 2.80t + 0.61t³

velocity is the derivative of position

v(t) = 2.80 + 1.83t²

acceleration is the derivative of velocity

a(t) = 3.66t

F = ma = 5.50(3.66(4.10)) = 82.533 N

which should be rounded to no more than three significant digits and arguably only two due to the 0.61 factor.

F = 82.5 N or 83 N

Yes the units are Newtons, cannot tell what your system will accept. May not want the units at all.