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Wewaii [24]
3 years ago
11

Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

1.1m/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 <180, 0, 360 > N.How much work does Jack do?WJack = JHow much work does Jill do?WJill = JWhat is the final speed of the boat?vf = m/s
Physics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

The workdone by Jack is  W_{jack} = -1050J

The workdone by Jill is  W_{Jill} = 0J

The final velocity is  v = 1.36 m/s

Explanation:

From the question we are given that

          The mass of the boat is m_b = 3300kg

          The initial position of the boat is   P_i  = (2 \r  i  + 0 \r j + 3\r k)m

           The Final position of the boat is  P_f = (4\r i + 0 \r j + 2\r k )\ m

           The Force exerted by Jack \r F = (-420\r i + 0 \r j + 210\r k) \ N

             The Force exerted by Jill  \r F_{Jill} =(180 \r i + 0\r j + 360\r k)

Now to obtain the displacement made we are to subtract the final position from the initial position

                                 \r P = P_f - P_i

                                    = (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k  )

                                     = (2\r i + 0\r j -\r k )m

Now that we have obtained the displacement we can obtain the Workdone

  which is mathematically represented as

                                                   W =\r  F * \r P

 The amount of workdone by jack would be

                                               W_{jack} =\r  F * \r P

                                                 = [(-420\r i +0\r j +210\r k)(2\r  i + 0\r j - \r k)]

                                                 = (-420) (2) + (210)(-1)

                                                = -840 - 210

                                               =-1050J

  The amount of workdone by Jill would be

                                                 W_{Jill} =\r  F * \r P

                                                        = [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]

                                                       = (180 )(2) +(360)(-1)

                                                       =0J

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]

             -1050  = 0.5*3300 [*v^2- (1.1)^2]

            -1050 = 1650 [v^2 -1.21]

               0.6363 = v^2 -1.21

                   v^2 = 0.6363+1.21

                    v^2 =1.846

                    v = 1.36\ m/s

                   

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