Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kingkos

  • 3 years ago

a force of 750n streches a certain spring a distance of 0.150 m.what is the pe of the spring when a 60kg mass hangs vertically from it.

  • This Question is Closed
  1. faheemullah
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    m= 60 kg F=750 N X=.150 m P.e=? as P.e=1/2 k\[x ^{2}\] but k=? for k use F=-kx K=-F/x=750/.150=-5000 put it in P.e equation i.e P.E=1/2k\[x ^{2}\]=0.5 x (-5000) x \[0.150^{2}\]=-56.25 joules

  2. 92champs
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Correct until final part. The x is not .15m with the 60kg mass. x=F/-k =mg/k =60g/5000 ==.11772 Therefore P.E.= .5 * 5000 * .11772^2=34.645J Note minus sign just to show force in opp. dir. to displacement

  3. faheemullah
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    k comes from ?

  4. 92champs
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    As you showed, F=-kx (Hooke's Law). The first part is in order to get k, which doesnt change for the second part as its the same spring

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.