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stira [4]
3 years ago
12

At a dock, a crane lifts a 2009 kg container 20 m, swings it out over the deck of a freighter, and lowers the container into the

hold of the freighter, which is 8 m below the level of the dock. The acceleration of gravity is 9.81 m/s 2 . Neglect friction losses. How much work is done by the crane on the container? Answer in units of kJ
Physics
1 answer:
deff fn [24]3 years ago
6 0

Answer:

W = 157.5kJ

Explanation:

Assuming it moves the container at constant speed, the work done by the crane will be equal to the variation of the potential gratitational energy on the container:

Wc = \Delta E = m*g*(h2 - h1)  where h2= -8m  and  h1=0m

Wc = 157.5kJ

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Answer with explanation:

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a.Initial velocity of plate,u_2=0

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Where g=9.8 m/s^2

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According to law of conservation of momentum  

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0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

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b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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