Answer
From $t=0$ to $t = 4.0~s$, the work done on the object by the force is $~~528~J$
Work Step by Step
$x = 3.0t-4.0t^2+1.0t^3$
$v = 3.0-8.0t+3.0t^2$
We can find the velocity at $t = 0$:
$v = 3.0-8.0t+3.0t^2$
$v = 3.0-8.0(0)+3.0(0)^2$
$v = 3.0~m/s$
We can find the kinetic energy at $t = 0$:
$K_i = \frac{1}{2}mv^2$
$K_i = \frac{1}{2}(3.0~kg)(3.0~m/s)^2$
$K_i = 13.5~J$
We can find the velocity at $t = 4.0~s$:
$v = 3.0-8.0t+3.0t^2$
$v = 3.0-8.0(4.0)+3.0(4.0)^2$
$v = 19.0~m/s$
We can find the kinetic energy at $t = 4.0~s$:
$K_f = \frac{1}{2}mv^2$
$K_f = \frac{1}{2}(3.0~kg)(19.0~m/s)^2$
$K_f = 541.5~J$
We can find the work done on the object by the force from $t=0$ to $t = 4.0~s$:
$K_i+W=K_f$
$W = K_f-K_i$
$W = 541.5~J-13.5~J$
$W = 528~J$
From $t=0$ to $t = 4.0~s$, the work done on the object by the force is $~~528~J$.
