Physics, 23.03.2020 23:02 angie249

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

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Answer from: elgallo8773

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

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