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

Answer Details

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