Which heights to use when in the equation

Question

When a 0.40-kg block is suspended on a vertical spring, it causes it to stretch 3.00 cm. If the block is now pulled 7.80 cm below its equilibrium position and released, what is the speed of the block when it is 4.30 cm below the equilibrium position?

Answer 1

There's an alternate way to find the speed without using calculus, but using conservation of energy. The maximum elastic potential energy occurs when the block is pulled 7.8 cm below the equilibrium point: initial EPE = 1/2 k y² = 1/2 (130.7) (0.078 m)² = 0.3976 J. At 4.3 cm below the equilibrium position, EPE = 1/2 (130.7) (0.043 m)² = 0.1208 J. The difference has gone into kinetic energy 1/2 m v² = 0.2768 J. Then v² = 2(0.2768) / (0.40 kg) = 1.384, and v = 1.18 m/s. Somehow this differs from Mark's answer.

Answer 2

First find the spring constant: F=ky where F = force, k = spring constant, y = vertical distance (+ is down)

The spring constant is k = F/y = (.4)(9.8)/(.03) = 130.67 kg/s²

Using F = ma = dv/dt where m = mass = .4 kg and a = dv/dt or dv/dy(dy/dt) = v(dv/dt)

when you add the forces, we get (.4 kg)(9.8 m/s²) - 130.67y

So,

(.4)v(dv/dt) = 3.92 - 130.67y or v(dv/dt) = 9.8 - 326.67y or v dv = (9.8 - 326.67y) dy

Which integrates to

(1/2)v² = 9.8y - (1/2)(326.67)y² + Constant

or v² = 19.6y - 326.67y² + K

then to find the constant, we know that velocity = 0 when y = .078

K = .459

So, when y = .043 v² = .698 or v = .835 m/s