A scale has a spring whose elasticity constant is N 1,000 If you place a mass m in it of 10 kg, how much will its spring stretch? Consider g = 10 m S

The formula for Hooke's law is: F = kx where F is the force exerted by the spring, k is the spring constant, and x is the stretch of the spring. In this case, the mass of the object is 10 kg, and the acceleration due to gravity is 10 m/s^2. Therefore, the weight of the object is: W = mg = 10 kg * 10 m/s^2 = 100 N The force exerted by the spring is equal to the weight of the object, so we can write: F = 100 N The spring constant is given as 1000 N/m, so we can write: k = 1000 N/m Substituting these values into the formula for Hooke's law, we get: 100 N = 1000 N/m * x Solving for x, we get: x = 0.1 m Therefore, the spring will stretch by 0.1 m when a mass of 10 kg is placed on it.