Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,401 questions

17,804 answers

1,438 comments

43,435 users

A spring with a force constant of 5.2 N/m

+1 vote
A spring with a force constant of 5.2 N/m has
a relaxed length of 2.64 m. When a mass is
attached to the end of the spring and allowed
to come to rest, the vertical length of the
spring is 3.66 m.
Calculate the elastic potential energy
stored in the spring.
Answer in units of J
asked Jan 27, 2013 in PHYSICS by anonymous Apprentice

1 Answer

+1 vote

Spring constant,k=5.2 N/m

Relaxed length,x0=2.64 m

Vertical length,x=3.66 m

--------------------------------

Stretched length,δ=x-x0

                                 = 3.66 -2.64

                                 =1.02 m

Potential energy

PE=(1/2)kδ^2

      =(1/2)*5.2*(1.02)^2

     =2.705 J

answered Jan 27, 2013 by bradely Mentor
is this right?

PE=(1/2)kδ^2

      =(1/2)*5.2 N/m*(1.02 m)^2

      = 2.70504 N. m

      = 2.70504 J

Yes, answer is right

Related questions

...