Answer:
The workdone by Jack is [tex]W_{jack} = -1050J[/tex]
The workdone by Jill is [tex]W_{Jill} = 0J[/tex]
The final velocity is [tex]v = 1.36 m/s[/tex]
Explanation:
From the question we are given that
The mass of the boat is [tex]m_b = 3300kg[/tex]
The initial position of the boat is [tex]P_i = (2 \r i + 0 \r j + 3\r k)m[/tex]
The Final position of the boat is [tex]P_f = (4\r i + 0 \r j + 2\r k )\ m[/tex]
The Force exerted by Jack [tex]\r F = (-420\r i + 0 \r j + 210\r k) \ N[/tex]
The Force exerted by Jill [tex]\r F_{Jill} =(180 \r i + 0\r j + 360\r k)[/tex]
Now to obtain the displacement made we are to subtract the final position from the initial position
[tex]\r P = P_f - P_i[/tex]
[tex]= (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k )[/tex]
[tex]= (2\r i + 0\r j -\r k )m[/tex]
Now that we have obtained the displacement we can obtain the Workdone
which is mathematically represented as
[tex]W =\r F * \r P[/tex]
The amount of workdone by jack would be
[tex]W_{jack} =\r F * \r P[/tex]
[tex]= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)][/tex]
[tex]= (-420) (2) + (210)(-1)[/tex]
[tex]= -840 - 210[/tex]
[tex]=-1050J[/tex]
The amount of workdone by Jill would be
[tex]W_{Jill} =\r F * \r P[/tex]
[tex]= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)][/tex]
[tex]= (180 )(2) +(360)(-1)[/tex]
[tex]=0J[/tex]
According to work energy theorem the Workdone is equal to the kinetic energy of the boat
[tex]W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2][/tex]
[tex]-1050 = 0.5*3300 [*v^2- (1.1)^2][/tex]
[tex]-1050 = 1650 [v^2 -1.21][/tex]
[tex]0.6363 = v^2 -1.21[/tex]
[tex]v^2 = 0.6363+1.21[/tex]
[tex]v^2 =1.846[/tex]
[tex]v = 1.36\ m/s[/tex]
