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jarptica [38.1K]

3 years ago

6

Superman needs to help a construction worker and lifts a 425 kg steel beam vertically upward a distance of 117 m. How much work

does Superman do on the beam if the beam accelerates upward at 1.8 m/s2? Neglect frictional forces. A. 4.0x105 J B. 3.4x105 J C. 5.8x105 J D. 4.9x105 J

1 answer:

s2008m [1.1K]3 years ago

8 0

Answer:

Work done = 5.8 × 10⁵ J (Approx)

Explanation:

Given:

Mass of the beam (m) = 425 kg

Distance (s) = 117 m

Acceleration (a) = 1.8 m/s²

Find:

Work done.

Computation:

⇒ Work done = Force × Displacement

⇒ Work done = Mass (Acceleration + Acceleration of gravity) × Displacement

⇒ Work done = 425 (1.8 + 9.81) × 117

⇒ Work done = 425 (11.61) × 117

⇒ Work done = 577,307.25

⇒ Work done = 5.8 × 10⁵ J (Approx)

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Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

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$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

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$q_{1} = q / 2$ .

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