Answered

Find accurate and reliable answers to your questions on IDNLearn.com. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.

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.

Sagot :

Eduard22slyidn

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

Data obtained from the question

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

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

How to determine the spring constant

  • 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

How to determine the work done

  • 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:

https://brainly.com/question/9199238

#SPJ1

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.