Prove that the stress needed to double the length of a wire of uniform cross section is equal to its Young's modulus. Assume that hooke's law is obeyed during the extension of the wire.
Answers
Answer:Where FFF is the force, xxx is the length of extension/compression and kkk is a constant of proportionality known as the spring constant which is usually given in \mathrm{N/m}N/m.
Though we have not explicitly established the direction of the force here, the negative sign is customarily added. This is to signify that the restoring force due to the spring is in the opposite direction to the force which caused the displacement. Pulling down on a spring will cause an extension of the spring downward, which will in turn result in an upward force due to the spring.
It is always important to make sure that the direction of the restoring force is specified consistently when approaching mechanics problems involving elasticity. For simple problems we can often interpret the extension xxx as a 1-dimensional vector; in this case the resulting force will also be a 1-dimensional vector and the negative sign in Hooke’s law will give the correct direction of the force.
When calculating xxx, it is important to remember that the spring itself will also have some nominal length L_0L
0
L, start subscript, 0, end subscript. The total length LLL of a spring under extension is equal to the nominal length plus the extension, L=L_0 + xL=L
0
+xL, equals, L, start subscript, 0, end subscript, plus, x. For a spring under compression, it would be L=L_0-xL=L
0
−xL, equals, L, start subscript, 0, end subscript, minus, x.