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3 years ago

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Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahea

d of her--five kilometers--and she's pulling with a force of 200 newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop

1 answer:

vitfil [10]3 years ago

5 0

Answer:

W = 819152 J = 819.15 KJ

Explanation:

The work done by Juri can be given by the following formula:

$W = FdCos\theta$

where,

W = Work done = ?

F = Force = 200 N

d = distance = 5 km = 5000 m

θ = angle to horizontal = 35°

Therefore,

W = (200 N)(5000 m)Cos 35°

<u>W = 819152 J = 819.15 KJ</u>

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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

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Answer:

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

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

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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 $P_f = (4\r i + 0 \r j + 2\r k )\ m$

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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

$\r P = P_f - P_i$

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

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$W =\r F * \r P$

The amount of workdone by jack would be

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

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The amount of workdone by Jill would be

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

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

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

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According to work energy theorem the Workdone is equal to the kinetic energy of the boat

$W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]$

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$0.6363 = v^2 -1.21$

$v^2 = 0.6363+1.21$

$v^2 =1.846$

$v = 1.36\ m/s$

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