A climber with a mass of 75 kg is attached to a 6.5 m rope with a cross sectional area of 3.5×10−4 m2. When the climber hangs from the rope, it stretches 3.4×10−3 m. What is the Young's modulus of the rope?

Please help me :)

Young's modulus is

E = stress / strain,
E= F•L / A₀•ΔL = m•g•L / A₀•ΔL=
=75•9.8•6.5/3.5•10−4•3.4•10−3 =
=4•10^9 Pa.

To calculate the Young's modulus of the rope, we need to use Hooke's law, which states that the tension in a stretched or compressed object is directly proportional to the strain applied to it.

The formula for Hooke's law is as follows:
Stress (σ) = Young's Modulus (Y) × Strain (ε)

In this case, the mass of the climber and the length and cross-sectional area of the rope are given. However, we need to calculate the stress and strain first.

1. Calculate the force exerted by the climber on the rope.
The force (F) is equal to the weight of the climber, which can be calculated using Newton's second law.
F = mass × gravity
F = 75 kg × 9.8 m/s^2 (acceleration due to gravity)
F = 735 N

2. Calculate the stress.
Stress (σ) is the force per unit area.
Stress (σ) = force (F) ÷ cross-sectional area (A)
Stress (σ) = 735 N ÷ (3.5 × 10^-4 m^2)
Stress (σ) = 2.1 × 10^6 N/m^2 (also known as Pascal or Pa)

3. Calculate the strain.
Strain (ε) is the change in length divided by the original length.
Strain (ε) = change in length (ΔL) ÷ original length (L₀)
Strain (ε) = 3.4 × 10^-3 m ÷ 6.5 m
Strain (ε) = 5.23 × 10^-4

4. Calculate Young's Modulus.
Plug the values of stress (σ) and strain (ε) into Hooke's law formula.
Young's Modulus (Y) = Stress (σ) ÷ Strain (ε)
Young's Modulus (Y) = (2.1 × 10^6 N/m^2) ÷ (5.23 × 10^-4)
Young's Modulus (Y) ≈ 4.014 × 10^9 N/m^2

Therefore, the Young's modulus of the rope is approximately 4.014 × 10^9 N/m^2.