## University Physics with Modern Physics (14th Edition)

$F = 122~N$
$y(t) = (2.80~m/s)~t+(0.610~m/s^3)~t^3$ $v(t) = \frac{dy}{dt} = (2.80~m/s)+(1.830~m/s^3)~t^2$ $a(t) = \frac{dv}{dt} = (3.660~m/s^3)~t$ 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: $\sum F = ma$ $F-mg = ma$ $F = m(g+a)$ $F = (5.00~kg)(9.80~m/s^2+14.64~m/s^2)$ $F = 122~N$