#### Answer

$F = 122~N$

#### Work Step by Step

$y(t) = (2.80~m/s)~t+(0.610~m/s^3)~t^3$
$v(t) = \frac{dy}{dt} = (2.80~m/s)+(1.830~m/s^3)~t^2$
$a(t) = \frac{dv}{dt} = (3.660~m/s^3)~t$
We can find the acceleration at t = 4.00 s:
$a = (3.660~m/s^3)~t$
$a = (3.660~m/s^3)(4.00~s)$
$a = 14.64~m/s^2$
We can use a force equation to find the magnitude of $F$ at t = 4.00 s:
$\sum F = ma$
$F-mg = ma$
$F = m(g+a)$
$F = (5.00~kg)(9.80~m/s^2+14.64~m/s^2)$
$F = 122~N$