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!
- Anonymous5 months agoFavorite 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)
- oubaasLv 75 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)
- Andrew SmithLv 75 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