Answer
$F=(-7.98N)i^{\wedge}$
Work Step by Step
It is given that $x_(t)=-13.00+2.00t+4.00t^2-3.00t^3$.
Take the derivative $\frac{d_x}{d_t}$, we obtain velocity:
$\frac{d_x}{d_t}=2.00+8.00t-9.00t^2$
Taking the second derivative, we obtain acceleration $a$:
$a=\frac{d^2_x}{dt^2}=8.00-18.00t$
Given that $t=3.40s$,
$a=8.00-18.00(3.40)$
$a=-53.2\frac{m}{s^2}$
We know that
$F=ma$
We plugin the known values to obtain:
$F=0.150(-53.2)$
$F=(-7.98N)i^{\wedge}$
