Show that work done in a wire, E=YAsqr(e)/2l, explainthe symbols. pls help.thanks
To show the work done in a wire using the given formula E = (YAsqr(e))/(2l), we need to understand the symbols involved and the variables they represent.
Symbols used:
- E: Represents the work done in the wire.
- Y: Represents Young's modulus, a measure of the stiffness of the material the wire is made of.
- A: Represents the cross-sectional area of the wire.
- e: Represents the elongation or change in length of the wire.
- l: Represents the original length of the wire.
Explanation of the formula:
The formula for the work done in a wire is derived from Hooke's Law, which states that the force required to extend or compress an elastic material is directly proportional to the amount of deformation, as long as the elastic limit of the material is not exceeded.
The elongation (e) of the wire is directly proportional to the force applied (F) and inversely proportional to Young's modulus (Y), as given by e = (Fl)/(YAsqr), where F is the force applied and l is the original length of the wire.
To find the work done (E), we need to calculate the force applied and the elongation of the wire. The force can be calculated using Hooke's Law as F = YAsqr(e)/l.
Now, to find the work done, we need to integrate the force over the elongation of the wire. The formula for work done is E = ∫(F * de), where ∫ represents integration.
By substituting the value of F from Hooke's Law, we get E = ∫((YAsqr(e)/l) * de). Simplifying this further, we get E = (YAsqr(e))/(2l), which is the formula for the work done in the wire.
So, in conclusion, the given formula E = (YAsqr(e))/(2l) represents the work done in a wire, where Y is Young's modulus, A is the cross-sectional area, e is the elongation, and l is the original length of the wire.