Question

Jack and Jill are maneuvering a 3400 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is 1.1 m/s. As the boat moves to position < 4, 0, 2 > m, Jack exerts a force of < -420, 0, 210 > N, and Jill exerts a force of < 160, 0, 320 > N. How much work does Jack do?How much work does Jill do? How much work does Jill do?What is the final speed of the boat?

Answer 1

displacement x = r2 - r1 = < ( 4-2) , (0-0) , (2 - 3) >

x = < 2 , 0 , -1 >

work done by Jack = F*x = < -420 , 0 , 210 > . < 2 , 0 , -1 >

WJack =   (-420*2) + (0*0) + (210*-1)

WJack = -1050 J

------------------------------

work done by Jill = F*x = < 160 , 0 , 320 > . < 2 , 0 , -1 >

WJack =   (160*2) + (0*0) + (320*-1)

WJack = 0 J

========================

from work energy relation

total work = change in KE

WJack + WJill = (1/2)*m*(vf^2 - vi^2)

-1050 + 0 = (1/2)*3400*(vf^2 - 1.1^2)

vf = 0.77 m/s <<<<----------answer