Answer:
ΔL = 0.2697 m = 26.97 cm
Explanation:
Given
∅ = 2.5 cm = 0.025 m
E = 10*10¹⁰ N/m²
L = 900 m
m = 1500 Kg
ΔL = ?
We get P = W as follows
W = m*g = (1500 Kg)(9.81 m/s²) = 14715 N
then we apply the equation of Hooke's Law
ΔL = P*L / (A*E)
⇒ ΔL = 14715 N*900 m / (π*0.25*(0.025 m)²*10*10¹⁰ N/m²)
⇒ ΔL = 0.2697 m = 26.97 cm
The change in the length of the cable when the load is attached to it is 26.9 cm.
Hooke's law is used to express the elastic nature of the metal wire. According to Hooke's law, the small increase in its length when stretched by an applied force doubles each time the force is doubled.
Given that Young's modulus Y of the cable is 10×10^10N/m2. The extended length L of the cable is 900 m and the mass m of ore is 1500 kg. The diameter d of the cable is 2.5 cm or 0.025 m.
The force applied by the gravity on the ore is calculated as given below.
[tex]F = mg[/tex]
[tex]F = 1500 \times 9.8[/tex]
[tex]F = 14700\;\rm N[/tex]
This force is equivalent to the work done in stretching the cable. So by Hooke's law, the change in the length of the cable is calculated as given below.
[tex]\Delta L = \dfrac {W L}{YA}[/tex]
Where [tex]\Delta L[/tex] is the change in length of the cable, W is the work done in stretching the cable, A is the cross-sectional area of the cable.
[tex]\Delta L = \dfrac {14700 \times 900}{10\times 10^{10} \times \pi \times (\dfrac {0.025}{2})^2}[/tex]
[tex]\Delta L = \dfrac {13230000}{49062500}[/tex]
[tex]\Delta L = 0.269 \;\rm m[/tex]
[tex]\Delta L = 26.9 \;\rm cm[/tex]
Hence we can conclude that the change in the length of the cable when the load is attached to it is 26.9 cm.
To know more about Hooke's law, follow the link given below.
https://brainly.com/question/3355345.
