Question

My answer was 19.8 cm but it was incorrect.

A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m.

If you suddenly put a 3.00 kg adobe brick in the basket, find the maximum distance that the spring will stretch. Answer: 3.92 cm

If, instead, you release the brick from 1.00 m above the basket, by how much will the spring stretch at its maximum elongation?

My answer was 19.8 cm but it was incorrect.

Answer 1

At equilibrium, the spring stretches an amount X given by
k X = W = M g
X = 3.00*9.81/1500 = 1.96 *10^-2 m = 1.96 cm
When the brick is dropped suddenly, it stretches twice as far and oscillates about the 1.96 cm deflection equilibrium position.
The maximum deflection is 2*1.96 = 3.92 cm. I assume you got that part right

In the second case, use conservation of energy. The brick falls a distance 1.00 + Xmax and converts gravitational P.E. = M g (1 + Xmax) into spring energy (1/2)k Xmax^2
29.4 (1 + Xmax) = 750 Xmax^2
Solve the quadratic equation.

Xmax = (1/1500)[29.4 + sqrt(864+88200)] = 0.219 m = 21.9 cm

You may have forgotten to include the additional PE loss during stretching