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 =

Explanation:

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

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

a and c

c is 100% but i am confused in a