In old science fiction movies, writers attempted to come up with novel ways of l

Question

In old science fiction movies, writers attempted to come up with novel ways of launching spacecraft toward the moon. In one hypothetical case, a screenwriter envisioned launching a moon probe from a deep, smooth tunnel, inclined at 73.0° above the horizontal. At the bottom of the tunnel a very stiff spring designed to launch the craft was anchored. The top of the spring, when the spring is unstressed, is 32.0 m from the upper end of the tunnel. The screenwriter knew from his research that to reach the moon, the 318-kg probe should have a speed of at least 11.2 km/s when it exits the tunnel. If the spring is compressed by 83.0 m just before launch, what is the minimum value for its force constant to achieve a successful launch? Neglect friction with the tunnel walls and floor.
kN/m

Explanation / Answer

Data: x = compressed length of spring before launching = 83 meters
m= mass of probe = 318 kg
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
h = vertical distance of tunnel exit from ground = (83 + 32) sin 73
                                                                                 = 109.9 m
V = exit velocity of probe in tunnel = 11.2 km/sec = 11,200 m/s

Solution: From the law of conservation of energy,
Spring energy = PE + KE of probe (at the tunnel exit)

Spring energy = (1/2)(k)x^2

Potential energy = mgh

Kinetic energy = (1/2)mV^2

where k = spring constant

Hence,
(1/2)(k)x^2 = mgh + (1/2)mV^2
k = 2m [ gh + (1/2)(V^2)] / x^2

k = [2 (318)/83^2][(9.8 * 109.9) + (1/2)(11200)^2]
= 5790478.3 N/m = 5790.478 kN/m Ans: Spring constant: k = 5790.478 kN/m (or) 5790478.3 N/m

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