Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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
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
k comes from ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question