A cable is 100 m long and has a cross-section area of 1 mm2. A 1000-N force is applied to stretch the cable. If Young’s modulus for the cable is 1.0x1011 N/m2
, how far does it stretch?
A. 10. m
B. 1.0 m
C. 0.10 m
D. 0.01 m
E. 0.001 m
1.0
To calculate the change in length (stretch) of the cable, we can use Hooke's Law, which states that the change in length is directly proportional to the force applied and inversely proportional to the cross-sectional area and Young's modulus.
The formula for Hooke's Law is:
ΔL = (F * L0) / (A * Y)
Where:
ΔL is the change in length
F is the force applied
L0 is the original length of the cable
A is the cross-sectional area
Y is Young's modulus
Given:
F = 1000 N
L0 = 100 m
A = 1 mm^2 = 0.000001 m^2
Y = 1.0x10^11 N/m^2
Plugging in the values into the formula:
ΔL = (1000 N * 100 m) / (0.000001 m^2 * 1.0x10^11 N/m^2)
ΔL = 1.0x10^8 m / 1.0x10^-5 m^2
ΔL = 1.0x10^13 m
The cable stretches by 1.0x10^13 meters.
Since none of the answer choices match this value, there seems to be an error in the question or in the given values. Please double-check the question or if any values were mistakenly entered.
To find out how far the cable stretches, we can use Hooke's Law, which states that the extension of an object is directly proportional to the force applied to it.
Hooke's Law can be expressed as:
F = k * ΔL
Where:
F is the force applied to the cable (1000 N in this case)
k is the spring constant or Young's modulus (1.0x1011 N/m2 in this case)
ΔL is the change in length or extension of the cable (what we want to find)
Rearranging the formula to solve for ΔL, we get:
ΔL = F / k
Substituting the given values:
ΔL = 1000 N / 1.0x1011 N/m2
Calculating this using a calculator, we get:
ΔL = 1.0x10^-8 m
Therefore, the cable stretches by 0.00000001 m.
Since this value is very small, it's often easier to express it in scientific notation, which gives us:
ΔL = 1.0x10^-8 m = 0.01 mm
So, the correct answer is D. 0.01 m.