A 970 kg motorboat accelerates away from a dock at 2.7 m/s2. Its propeller provides a thrust force of 3.9 kN. What is the drag force exerted by the water on the boat?

Question

anonymous · Answer

F=ma
m=970kg
a=2.7 m/s^2
F=(970)(2.7)
F=2619N
Since the motorboat is accelerating AWAY from the dock, the thrust force will be negative. 
*Don't forget to convert 3.9kN to N
Drag force=F_1+F_2=-3900N+2619N
Answer: -1281N or -1.218kN

anonymous · Answer

I too agree with the above solution...its explicitly based upon the newtons dynamic equation..net force=mass times acceleration..
or thrust force-drag force (here viscous force)=mass times acceleration.

theraggedydoctor · Answer

;PPP