Question

a spring has a spring constant of 0.5n/m. how far has it stretched if it stores 5j of elastic potential energy? Show working out please.

Answer 1

4.47 m

Step-by-step explanation:

Elastic potential energy is given by 1/2 k x^2

k = spring constant = .5 N/m

and x is the amount of stretch or compression

5 j = 1/2 ( .5 N/m ) x^2

5 / ( 1/2 *.5) = x^2

x = 4.47 m

Answer 2

Spring extension:

Δx ≈ 4.5 m

Step-by-step explanation:

Given:

k = 0.5 N/m

E = 5 J

__________

Δx - ?

Potential energy of a compressed (stretched) spring:

E = k·(Δx)² / 2

Spring extension:

Δx = √ (2·E / k)

Δx = √ (2·5 / 0.5 ) ≈ 4.5 m