When a wire of lenght 5m & radius 0.5mm is stretched by a load of 49N,the elongation produced in the wire is 0.1cm. Find the energy stored per unit volume of the wire.
σ=F/A=F/πr²
ε=ΔL/L
Hook’s law for the wire:
k ε= σ
k= σ/ ε= F L /πr²ΔL
E=k•ΔL²/2= F• L•ΔL²/2πr²ΔL=
=F•L•ΔL/2πr²=49•5•10⁻³/2•π•25•10⁻⁸=1.56•10⁵ J
E₀=E/V= E/πr²L = 1.56•10⁵/π•25•10⁻⁸•5=3.97•10¹⁰ J/m³
To find the energy stored per unit volume of the wire, we need to determine the strain energy stored in the wire and then divide it by the volume.
First, let's calculate the initial volume of the wire. The wire can be approximated as a cylinder, so the initial volume (V) is given by:
V = π * r^2 * h
where r is the initial radius of the wire and h is the initial length of the wire.
Given:
Initial length (h) = 5m
Initial radius (r) = 0.5mm = 0.5 * 10^(-3)m
Plugging in these values, we get:
V = π * (0.5 * 10^(-3))^2 * 5
V = π * (0.5^2 * 10^(-6)) * 5
V = π * 0.25 * 10^(-6) * 5
V = 3.14 * 0.25 * 10^(-6) * 5
V = 3.14 * 0.25 * 5 * 10^(-6)
V = 3.14 * 0.125 * 10^(-5)
V = 3.925 * 10^(-5) m^3 (approximately)
Next, we can calculate the strain energy stored in the wire.
Strain energy (U) can be given by:
U = (1/2) * F * Δl
where F is the force applied to the wire and Δl is the elongation produced in the wire.
Given:
Force (F) = 49N
Elongation (Δl) = 0.1cm = 0.1 * 10^(-2)m
Plugging in these values, we get:
U = (1/2) * 49 * 0.1 * 10^(-2)
U = 0.5 * 49 * 0.1 * 10^(-2)
U = 0.5 * 4.9 * 10^(-3)
U = 2.45 * 10^(-3) J
Finally, to find the energy stored per unit volume, we just need to divide the strain energy by the volume:
Energy stored per unit volume (E) = U / V
E = (2.45 * 10^(-3)) / (3.925 * 10^(-5))
E = 6.24 J/m^3 (approximately)
Therefore, the energy stored per unit volume of the wire is approximately 6.24 Joules per cubic meter.