when awire of length 5m and radius 0.5mm is stretched by a load of 49Nthe elongation produced in the wire is 0.1cm. find the energy stored per unit volum of the wire
To find the energy stored per unit volume of the wire, we need to calculate the strain energy (work done) in the wire and divide it by the volume.
First, let's calculate the cross-sectional area of the wire:
The radius of the wire is given as 0.5 mm, which is equal to 0.0005 m (since 1 mm = 0.001 m).
The cross-sectional area (A) of a wire is given by the formula: A = πr^2, where r is the radius.
Plugging in the values, we get:
A = π(0.0005)^2
A = π(0.00000025)
A ≈ 0.0000007854 m^2
Next, let's calculate the volume of the wire:
The length of the wire is given as 5 m.
The volume (V) of a wire is given by the formula: V = AL, where A is the cross-sectional area and L is the length.
Plugging in the values, we get:
V = 0.0000007854 × 5
V = 0.000003927 m^3
Now, let's calculate the strain energy (work done) in the wire:
The strain energy (U) in a wire is given by the formula: U = (1/2) × F × e, where F is the applied force and e is the elongation.
Plugging in the values, we get:
U = (1/2) × 49 × 0.001
U = 0.0245 J
Finally, let's calculate the energy stored per unit volume:
The energy stored per unit volume (E) is given by the formula: E = U / V, where U is the strain energy and V is the volume.
Plugging in the values, we get:
E = 0.0245 / 0.000003927
E ≈ 6234.628 J/m^3
Therefore, the energy stored per unit volume of the wire is approximately 6234.628 J/m^3.