A 5.00 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.

(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?

122.2

Explanation:

From the question, the equation of the force acting on the crate is,

F-mg = ma

F = ma+mg Equation 1

Where F = upward force, m = mass of the crate, a = acceleration of the crate, g = acceleration due to gravity of the crate.

Given: m = 5.0 kg, g = 9.8 m/s²

And a,

a = d²y/dt²

Where,

y(t) = 2.8t+0.61t³

d²y/dt² = 3.66t equation 2

when  t = 4.0 s.

substitute into equation 2

d²y/dt² = a = 3.66(4)

a = 14.64 m/s²

substitute into equation 1,

F = 5(14.64)+5(9.8)

F = 73.2+49

F = 122.2 N.

answer: its c lololo

0.55x745.7j and a calculator