. A copper wire 1.00 mm in diameter is used to support a mass of 6.0 kg. If the wire was stretched by 2.00 mm, what was the original length of the wire?
Y = stress/ strain r= 0.5 mm =5 X 10-4 m e= 2mm or 2 X 10-3
Given :
m = 6 kg F = 60 N
A =π r2 = ( 3.14)( 5 X 10-4) (5 X 10-4) = 78.5 X 10-8 = 7.85 X 10-7 m2
11.6 x 10 10 N/m 2 = 60N/ 7.85 X 10-7 m2
( 2 x 10 -3m)/ L0
11.6 x 10 10 N/m ( 2 x 10 -3m)/ L0 = 60N/ 7.85 X 10-7 m2
L0= 182 m
To solve this problem, we can use the formula Y = stress/strain. In this case, stress is the force applied to the wire (F) divided by the area of the wire (A), and strain is the change in length of the wire (e) divided by the original length of the wire (L0).
First, let's calculate the stress. We are given the force applied to the wire (F = 60 N) and the area of the wire (A = 7.85 x 10^-7 m^2). The stress is given by stress = F/A.
Stress = 60 N / 7.85 x 10^-7 m^2 = 7.64 x 10^7 N/m^2
Next, let's use the formula Y = stress/strain and rearrange it to solve for the original length of the wire (L0).
Y = stress/strain
11.6 x 10^10 N/m^2 = (7.64 x 10^7 N/m^2) / (2 x 10^-3 m / L0)
Simplifying, we can cross-multiply and solve for L0:
(11.6 x 10^10 N/m^2) * (2 x 10^-3 m / L0) = 7.64 x 10^7 N/m^2
(2.32 x 10^8 N/m) / L0 = 7.64 x 10^7 N/m^2
L0 = (2.32 x 10^8 N/m) / (7.64 x 10^7 N/m^2)
L0 = 182 m
Therefore, the original length of the wire was 182 meters.