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2 years ago

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A 183 lb man sits in the middle of a 94 lb, 14 ft long boat. The boat’s prow touches the pier, but the boat isn’t tied to it. Th

e man stands up and walks towards the pier. Assume there is negligible resistance between the boat and water, and that the boat’s center of mass is in the middle of the boat. By the time he reaches the boat’s prow, what is the distance between the prow and the pier?

1 answer:

Marina86 [1]2 years ago

5 0

Answer:

d = 9.25 ft

Explanation:

As we know that there is no frictional force as well as no other resistive external force on boat + man system

So here the center of mass of the system of boat + man will remain at same position

So we have

$m_1(L - x) + m_2(-x) = 0$

$m_1 = 183 lb$

$m_2 = 94 lb$

$L = 14 ft$

so we will have

$183(14 - x) - 94(x) = 0$

$183 \times 14 = 277 x$

$x = \frac{183\times 14}{277}$

$x = 9.25 ft$

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(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

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Negative sign shows the direction of motion of the object after collision is toward left.

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Using formula of kinetic energy

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Put the value into the formula

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Put the value into the formula

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$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

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(b). The mechanical energy dissipates during this play is 226.98 J.

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Answer:

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before it hits the ground = 46 % of entire distance

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so take t = 3.771 s

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h = 69.68 m

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