Answer:
The workdone by Jack is 
The workdone by Jill is 
The final velocity is 
Explanation:
From the question we are given that
The mass of the boat is 
The initial position of the boat is 
The Final position of the boat is 
The Force exerted by Jack 
The Force exerted by Jill 
Now to obtain the displacement made we are to subtract the final position from the initial position


Now that we have obtained the displacement we can obtain the Workdone
which is mathematically represented as
The amount of workdone by jack would be

![= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)]](https://tex.z-dn.net/?f=%3D%20%5B%28-420%5Cr%20i%20%2B0%5Cr%20j%20%2B210%5Cr%20k%29%282%5Cr%20%20i%20%2B%200%5Cr%20j%20-%20%5Cr%20k%29%5D)



The amount of workdone by Jill would be

![= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]](https://tex.z-dn.net/?f=%3D%20%5B%28180%20%5Cr%20i%20%2B%200%5Cr%20j%20%2B%20360%5Cr%20k%29%282%5Cr%20i%20%2B0%5Cr%20j%20-%5Cr%20k%29%5D)


According to work energy theorem the Workdone is equal to the kinetic energy of the boat
![W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]](https://tex.z-dn.net/?f=W%20%3D%20K.E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%2A%5Bv%5E2%20-%20%281.1%29%5E2%5D)
![-1050 = 0.5*3300 [*v^2- (1.1)^2]](https://tex.z-dn.net/?f=-1050%20%20%3D%200.5%2A3300%20%5B%2Av%5E2-%20%281.1%29%5E2%5D)
![-1050 = 1650 [v^2 -1.21]](https://tex.z-dn.net/?f=-1050%20%3D%201650%20%5Bv%5E2%20-1.21%5D)



