Question

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

Answer 1

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

Comments

is this right?

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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