# Amy hangs a weight on a newtonmeter. The energy stored in the spring is 0.045J She then increases the weight by 2N?

While spring constant = 400N/m

Calculate the total extension?

Any help would be really appreciated, thanks!

### 3 Answers

- Anonymous7 months agoFavourite answer
Key formulae: F = kx and E = ½kx²

k = 400N/m

The energy initially stored in the spring is E₁ = 0.045J. So if the initial extension is x₁:

E₁ = ½kx₁²

0.045 = ½ * 400 * x₁²

x₁ = √(2* 0.045/400) = 0.015m

She then increases the weight by 2N. ΔF = 2N.

Since x is proportional to F, increase in extension, Δx, is given by:

ΔF = kΔx

2 = 400Δx

Δx =2/400 = 0.005 m

Total extension = x₁ + Δx = 0.015 + 0.005 = 0.02m (=2cm)

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- oubaasLv 77 months ago
400/2*x^2 = 0.045

x = √0.090/400 = 0.0150 m (1.5 cm)

F = k*x = 0.015*400 = 6.0 N

x' = (F+2)/k = 8/400 = 0.02 m (2.0 cm)

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- Andrew SmithLv 77 months ago
If the spring constant is 400 N/m then the spring will be extended BY 1/200 m ( 0.5 cm)

Now if the energy is 0.045 J

and E = 1/2 k x^2 x = sqrt( 2E/k) = sqrt( 2*0.045/400) = 0.015 m

Add the two to get a total extension of 0.020 m

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