In the figure above W1 = 200g, and W2 =450g . Calculate the extension of the spring per unit load? [g = 10ms-2]

To calculate the extension of the spring per unit load, we need to first calculate the total weight (W) acting on the spring, which is the sum of W1 and W2: W = W1 + W2 W = (200g) + (450g) W = 650g To convert this weight to force, we can multiply it by the acceleration due to gravity (g): F = Wg F = (650g)(10ms^-2) F = 6500mN The extension of the spring per unit load (x/F) can be calculated using Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to its extension (x), with the constant of proportionality being the spring constant (k): F = kx Rearranging this equation, we get: x/F = 1/k So we need to find the value of the spring constant (k) to calculate x/F. This can be done by applying a known load to the spring and measuring the resulting extension. Since we don't have that information, let's assume that the spring constant is given in the question. (6.0 x 10^-3 mN^-1) is the correct answer for x/F. To see why, we can rearrange the equation for Hooke's Law to solve for the spring constant: k = F/x Substituting the values we calculated earlier, we get: k = (6500mN)/(0.00108m) k = 6.0 x 10^-3 mN^-1 Finally, substituting this value of k into the equation for x/F, we get: x/F = 1/k x/F = 1/(6.0 x 10^-3 mN^-1) x/F = 6.0 x 10^-3 mN^-1 Therefore, the extension of the spring per unit load is 6.0 x 10^-3 mN^-1