CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 10 (Capstone), Exercise 1 Solution
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A Redesigned Piston: Force, Acceleration, and Launch Speed

PROBLEM
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A redesigned piston has area A = 0.06 m^2 and the same steam
pressure, P = 200,000 Pa, over the same 3 m stroke, launching a
lighter 350 kg car. Find the new launch force, acceleration, and
launch speed.

SOLUTION
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Step 1: Find the new launch force.

    F = P A
    F = 200,000 x 0.06
    F = 12,000 N

Step 2: Find the new acceleration (Chapter 2, F = ma).

    a = F / m
    a = 12,000 / 350
    a = 34.29 m/s^2 (approximately)

Step 3: Find the launch speed (Chapter 1, v^2 = u^2 + 2as, u = 0).

    v^2 = 2 a d
    v^2 = 2 x 34.29 x 3
    v^2 = 205.7
    v = sqrt(205.7)
    v = 14.34 m/s (approximately)

ANSWER: The new launch force is 12,000 N, the acceleration is
approximately 34.29 m/s^2, and the launch speed is approximately
14.34 m/s.

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WHY THIS WORKS AS AN ANSWER

A larger piston area (0.06 m^2 vs. the capstone's original 0.05 m^2)
increases the launch force even at the same steam pressure, since
force scales directly with area (F = PA). Combined with a LIGHTER
car (350 kg vs. 400 kg), the resulting acceleration increases even
more sharply, since acceleration depends on both a larger force AND
a smaller mass (a = F/m) - both changes push acceleration, and
therefore launch speed, higher than the capstone's own original
12.25 m/s result. This exercise mirrors the exact same three-step
chain used in the capstone itself (Step 4: force, then acceleration;
Step 5: launch speed), just with different design parameters.