Potential Energy and Conservation of Energy Question 2.?
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.
- Modus OperandiLv 61 decade agoFavorite Answer
F = ma = 2 * 9.8 = 19.6
F(spring) = k*(distance compressed)
19.6 = 1960 * x
X = 19.6 / 1960
- SteveLv 71 decade ago
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