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Question

A spring has a spring constant of 4.4 x 10⁴ N/m. What is the change in elastic potential energy stored in the spring when its stretch is increased from 12.5 cm to 15 cm?

Answer 1

Potential energy of the spring

U = 1/2 kx²,

here k =4.4 x 10⁴ N/m - spring constant,

x - deformation of the spring.

To find the change in potential energy ΔU, you just need to subtract the values of 2 energies.

Remember that deformation should be converted to units - meters.

So, x₁ =12.5 x 10^-2 m and x₂ = 15 x 10^-2 m

Now we can write

ΔU = U₂ - U₁ = k/2 (x₂² - x₁²)

Putting the values we will have

ΔU = 1/2·4.4 x 10⁴ N/m ((15 x 10^-2m)² - (12.5 x 10^-2 m)²)

Finally we get ΔU = 151 J