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3 years ago

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Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that l

oad in 1 6 the time it takes the other. How much more power does the faster crane have?

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

Explanation:

Given

Load $W=20000\ N$

height to which load is raised $h=250\ m$

Another crane take $\frac{1}{6}$ th time to lift the load

Energy required required to lift the Weight

$E=W\times h$

$E=20000\times 250$

$E=5,000,000\ J$

Suppose $P_1$ and $P_2$ is the Power required to lift the weight in t and $\frac{t}{6}$ time

$E=P_1\times t$

$E=P_2\times \frac{t}{6}$

$P_1\times t=P_2\times \frac{t}{6}$

thus

$P_2=6P_1$

Second Crane requires 6 times more power than the slow crane

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(a) v = 1.54m/s, (b) Θ = 80.6° and (c) E = 52.92J

Explanation:

(a) <u>At the lowest point, the total energy is giving by:</u>

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<em>where, $E_{K}$ : kinetic energy, $E_{P}$ : potential energy, m: pendulum's mass, v: speed of the pendulum, g:gravity and y: heigh of the pendulum. </em>

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<u>When the string is at 64° from the vertical, the total energy is:</u>

$E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$ (2)

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$y_{f} = L - Lcos(\Theta)$ (3)

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$y_{f} = 4.3 m (1 -cos(64)) = 2.42 m$

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$E_{T}_{0} = E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$

$v_{f} = \sqrt \frac{2(E_{T}_{0} - mgy_{f})}{mg}}$

$v_{f} = \sqrt \frac{2(52.92 - 1.5 \cdot 9.8 \cdot 2.42)}{1.5 \cdot 9.8}} = 1.54 \frac{m}{s}$

(b) <u>Smilarly, by conservation of energy we can find the greatest angle, assuming that vf = 0 at the greatest angle reached</u>:

$\frac {1}{2} mv_{0}^{2} + mgy_{0} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$

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Using the heigh calculated in equation (3) we can find the angle:

$\Theta = Arccos (\frac {L - y_{f}}{L}) = Arccos (\frac {4.3 - 3.6}{4.3}) = 80.6^\circ$

(c) The total mechanical energy at the lowest point is giving by:

$E_{T} = E_{K} + E_{P} = \frac {1}{2} mv_{0}^{2} + 0 = \frac {1}{2} (1.5)(8.4)^{2} = 52.92 J$

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