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sashaice [31]
1 year ago
7

A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

done in stretching the spring from its natural length to 0.7 feet beyond its natural length? answer exactly or round to 2 decimal places.
Mathematics
1 answer:
ioda1 year ago
5 0

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Work done (Wd) =?

<h3>How to determine the spring constant</h3>
  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>
  • Spring constant (K) = 5 pound/foot
  • Extention (e) = 0.7 feet
  • Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

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Gala2k [10]

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Step-by-step explanation:

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aksik [14]

Using the surface area formula for rectangular and triangular prism, the surface area of the composite figure is: 444 m².

<h3>What is the Surface Area of the Composite Figure?</h3>

Total surface area = surface area of the top triangular prism + surface area of the bottom rectangular prism - area of the surface both share together.

Surface area of the top triangular prism = (S1 + S2+ S3)L + bh = (10 + 10 + 16)5 + (16)(6) = 276 m².

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