Question

A single force acts on a 3.0 kg particle-like object whose position
is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in
seconds. Find the work done by the force from t =0 to t = 4.0 s.

Answer 1

Work is defined as force multiply by distance. 
Work from t=0 to t=4 is W = W(4)-W(0) 

Here, the force is function of time. F = m*a. 
a can be obtained from second derivative of x. 
we will get a= 6*t - 8 

Then we can say F=3*(6t-8) = 18t -24 

W=F*x 
So, work in the function of time is: 

W(t)= (18t -24)*(3.0t-4.0t^2+1.0t^3) 

Work done from t=0 to t=4 is 

W = W(4)-W(0) 
W= 576 - 0 
W= 576 Joule 

I hope it helps