Potential Energy and Conservation of Energy Question 2.?

Question

A 2.0kg block is dropped from a height of 40 cm onto a spring of spring constant k = 1960 N/m. Find the maximum distance the spring is compressed.

Answer 1

F = ma = 2 * 9.8 = 19.6

F(spring) = k*(distance compressed)

19.6 = 1960 * x

X = 19.6 / 1960

Answer 2

The PE lost by the block when it stops is mg(.40 + Δ)
The spring when fully deflected will store this energy AND support the weight of the block. So,

.5kΔ² = mg(.4 + Δ) → Δ = .03254 m

y(due to weight) = W/k = mg/k = .01 m

Total deflection = y + Δ = .04254 m