Jack and Jill are maneuvering a 3100 kg boat near a dock. Initially the boat's position is lesserthan 2, 0, 3 greaterthan m and its speed is 1.9 m/s. As the boat moves to position lesserthan 8, 0, 0 greaterthan m, Jack exerts a force of lesserthan - 420, 0, 210 greaterthan N, and Jill exerts a force of lesserthan 180, 0, 360 greaterthan N. a. How much work does Jack do? W_Jack = b. How much work does Jill do? W_Jill =

The mass of the boat is 3100kg.

Initial position of boat is r1 = (2,0,3) m

And initial velocity of boat is 1.9m/s

New position r2 = (8,0,0) m

Jill Force = (180,0,360) N

Jack force = (-420,0,210) N

a. Jack work

Work is given as the dot product of force and displacement

W=F•d

The displacement is ∆r

∆r=r2-r1

∆r = (8,0,0) - (2,0,3)

∆r = (6,0,-3)

The force exerted by Jack is Fj = (-420,0,210)

Then,

W=F•∆r

W = (-420 i + 0 j + 210 k) • (6 i + 0j - 3 k)

Then, i. i=j. j=k. k=1

Also, i. j=j. i=k. i=i. k=j. k=k. j=0

Using this principle

Then,

W = (-420 * 6) + (0*0) + (210*-3)

W=-2520-630

W=3150J.

The workdone by Jack is 3150J

2. Work done by jill

Using the same principle and they have the same displacement

∆r = (6,0,-3)

Therefore

W=F•∆r

W = (180 i + 0 j + 360 k) • (6 i + 0j - 3 k)

Then, i. i=j. j=k. k=1

Also, i. j=j. i=k. i=i. k=j. k=k. j=0

Using this principle

Then,

W = (180 * 6) + (0*0) + (360*-3)

W=1080-1080

W=0J.

The workdone by Jill is 0J, she did no work

This show that Jill apply her force perpendicular to the displacement.

Because W=0 if and only if the displacement is zero or it is perpendicular to the applied force

3. I think we can still calculate the final velocity of the boat since we are given the mass

Using conservation of energy principle

∆K. E=W

K. E=1/2mv²

Total work is 0+3150=3150J

Then,

0.5M (Vf²-Vi²) = W

(Vf²-Vi²) = W/0.5M

Given than M=3100kg and Vi=1.9m/s

Vf²-Vi²=3150 / (0.5*3100)

Vf²-1.9²=2.03

Vf²=2.03+1.9²

Vf²=5.6423

Vf=√5.6423

Vf=2.38m/s.

A) WJack = - 3150 J

B) WJill = 0 J

Explanation:

M = 3100kg and initial speed (u) = 1.9m/s

Now, we know that work done = Force x displacement.

So in this question,

A) Work done by Jack (W) = F x Δr

From the question, force applied by Jack equals (-420, 0, 210) N

Also, since the boat moves from initial position of (2, 0, 3) m to final position of (8, 0, 0) m, thus the displacement (Δr) = (8, 0, 0) m - (2, 0, 3) m = (6,0, - 3) m

Thus work done by Jack (W) =

(-420, 0, 210) N x (6,0, - 3) m=

( - 420 x 6) + (0) + (210 x (-3)) =

- 3150J

B) Force applied by Jill = (180, 0, 360) N

Using the same principle, work done by Jill = (180, 0, 360) N x (6,0, - 3) m =

(180 x 6) + (0) + (360 x (-3)) = 0J