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Piston with linear spring, air supply (5 kg)

Short answer

Force balance includes piston weight, atmosphere, and linear spring with zero force at bottom. Area = 0.007854 m², weight pressure = 6.245 kPa. Initial spring pressure = 293.75 kPa, height = 0.05093 m. For 2 cm rise, extra spring pressure = 115.4 kPa. New pressure = 515.4 kPa.

A 5 kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa. The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown in figure below, the pressure is 400 kPa with volume of 0.4 L. The valve is opened to let some air in, causing the piston to rise 2 cm. Find the new pressure. [Figure: vertical cylinder containing air, closed by a piston held down by a linear spring above it; an air supply line with a valve connects to the cylinder.]

Given: m = 5 kg, D = 100 mm, P_atm = 100 kPa, P₁ = 400 kPa, V₁ = 0.4 L = 0.0004 m³, piston rises Δx = 2 cm. Take g = 9.81 m/s².

Idea: The spring force is zero when the piston is at the bottom, so the spring force is F = kx, where x = V/A is the height of the piston above the bottom. Force balance on the piston: P·A = P_atm·A + m g + k x

Step 1: Area A = (π/4)(0.1)² = 0.007854 m²

Step 2: Piston weight pressure m g / A = (5 × 9.81)/0.007854 = 6245 Pa = 6.245 kPa

Step 3: Spring contribution at state 1 k x₁ / A = P₁ − P_atm − m g/A = 400 − 100 − 6.245 = 293.75 kPa

Step 4: Initial height x₁ = V₁/A = 0.0004/0.007854 = 0.05093 m

Step 5: Spring constant k = (293.75 × 1000 × 0.007854)/0.05093 = 4.53 × 10⁴ N/m ≈ 45.3 kN/m

Step 6: New state (x₂ = x₁ + 0.02 = 0.07093 m) Extra spring pressure = k Δx / A = (45 300 × 0.02)/0.007854 = 115.4 kPa

P₂ = 400 + 115.4 = 515.4 kPa

Answer: New pressure ≈ 515 kPa

✓ Answer checked by Claude Opus 5.5 (high effort)

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