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Piston with linear spring, air supply (15 kg)

Short answer

Force balance P = P_atm + mg/A + kx/A. Using initial state, spring constant 2993 N/m. A 5 cm rise adds 8.47 kPa, giving new pressure 308.5 kPa.

A 15 kg piston in a cylinder with diameter of 0.15 m is loaded with a linear spring and the outside atmospheric pressure of 100 kPa, as shown in figure below. The spring exerts no force on the piston when it is at the lower position of the cylinder and for the state shown, the pressure is 300 kPa with volume of 0.02 m³. The valve is opened to let some air in, causing the piston to rise 5 cm. Find the new pressure. [Take g = 9.81 m/s²] [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 = 15 kg, D = 0.15 m, P_atm = 100 kPa, P₁ = 300 kPa, V₁ = 0.02 m³, rise Δx = 5 cm = 0.05 m, g = 9.81 m/s².

Force balance: P A = P_atm A + m g + k x, with spring force zero at the bottom (x = V/A).

Step 1: Area A = (π/4)(0.15)² = 0.017671 m²

Step 2: Piston weight pressure m g/A = (15 × 9.81)/0.017671 = 8327 Pa = 8.327 kPa

Step 3: Spring pressure at state 1 k x₁/A = 300 − 100 − 8.327 = 191.67 kPa

Step 4: Piston height x₁ = V₁/A = 0.02/0.017671 = 1.1318 m

Step 5: Spring constant k = (191.67 × 1000 × 0.017671)/1.1318 = 2993 N/m

Step 6: Increase in pressure for 5 cm rise ΔP = k Δx / A = (2993 × 0.05)/0.017671 = 8468 Pa = 8.47 kPa

P₂ = 300 + 8.47 = 308.5 kPa

Answer: New pressure ≈ 308.5 kPa

✓ Answer checked by Claude Opus 5.5 (high effort)

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