If a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? {g=10ms-2}

The force constant of a cord represents the force required to stretch it by a unit length. In this case, the cord stretches by 1.2 cm when a load of 1 kg is applied to it. We can use Hooke's Law to find the force constant of the cord. Hooke's Law states that the force required to stretch a spring or cord is proportional to the amount of stretch. Mathematically, it can be expressed as F = kx, where F is the force, k is the force constant, and x is the amount of stretch. In this case, we know that the load applied is 1 kg or 1*10^3 g. We can convert this to mass in kg by dividing it by the acceleration due to gravity, g, which is given as 10 m/s^2. Therefore, the mass of the load is: m = 1*10^3 g / (10 m/s^2) = 100 g = 0.1 kg The amount of stretch, x, is given as 1.2 cm or 0.012 m. Using Hooke's Law, we can write: F = kx The force, F, required to stretch the cord by 0.012 m is equal to the weight of the load, which is: F = mg = 0.1 kg * 10 m/s^2 = 1 N Therefore, we can solve for the force constant, k, as: k = F/x = 1 N / 0.012 m = 83.3 N/m Thus, the force constant of the cord is 83.3 N/m. Looking at the answer options, we can see that option B is the closest to our answer of 83.3 N/m. Therefore, the correct answer is option B, 833 Nm^-1.