How much work is needed to stretch this spring a distance of 5. 0 cm , starting with it unstretched?
1 answer:
The work done by the unstretched spring is 5 J.
<h3>What is the Hooke's law?</h3>Hooke's law states that the extension of a given material is directly proportional to the force applied as log as the elastic limit is not exceeded. Thus we have to know that; F = Ke
F = force applied
K = force constant
e = extension
Using the graph;
K = F/e
F = 200N
e = 5 cm or 5 * 10^-2 m
K = 200N/ 5 * 10^-2 m
K = 4000 N/m
Now;
Work = 1/2Ke^2
Work = 0.5 * 4000 N/m * (5 * 10^-2 m)^2
Work = 5 J
Learn more about Hooke's law:brainly.com/question/13348278
#SPJ1
You might be interested in
Answer:
a) E = ρ / e0
b) E = ρ*a / (e0 * r)
c) E = 0
Explanation:
Because of the geometry, the electric field lines will all have a radial direction.
Using Gauss law
Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.
We can make this cylinder of different radii to evaluate the electric field at different points.
Then:
A = 2*π*r (area of cylinder per unit of length)
Q/e0 = 2*π*r*E
E = Q / (2*π*e0*r)
Where Q is the charge contained inside the cylinder.
Inside the cable core:
There is a uniform charge density ρ
Q(r) = ρ * 2*π*r
Then
E = ρ * 2*π*r / (2*π*e0*r)
E = ρ / e0 (electric field is constant inside the charged cylinder.
Between ther inner cilinder and the tube:
Q = ρ * 2*π*a
E = ρ * 2*π*a / (2*π*e0*r)
E = ρ*a / (e0 * r)
Outside the tube, the charges of the core cancel each other.
E=0