HW Helper PF Gold P: 1,198 One way to do this question. At time 't' what is the mass of water left in the bucket? It is 40 - 0.2t lb The mass of the bucket is 4lb. So the total mass is 40 -0.2t + 4 = 44 - 0.2t lb. now, by newton's second law, d(mv)/dt=F(net). Therefore, mdv/dt+vdm/dt=F(net) Now, dv/dt=0 so, vdm/dt=(T-mg) where T is the force you apply to pull the bucket as a function of time. So T= vdm/dt+mg. T=(2)(-0.2) + (44-0.2t)(g) {Use the value of g in appropriate units} and time taken to reach the top is h/v=(80)/(2)=40 So work done is $$\int T.dx$$ Which is $$\int T.(vdt)$$ = $$\int T.(2)(dt)$$ from t=0 to t=40