Physics problem help --- finding force constant given angle, distances, weight and speed?

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 60.0° 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 31.0 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 95.0 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

Answer 1

Apply the law of conservation of energy to this problem.

Spring energy = potential energy of probe + kinetic energy 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

x = compressed length of spring before launching = 91 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 = (95 +31)sin 60

V = exit velocity of probe in tunnel = 11.2 km/sec = 11,200 m/sec.

Since

Spring energy = potential energy + kinetic energy

(1/2)(k)x^2 = mgh + (1/2)mV^2

Simplifying the above (and solving for "k"),

k = 2m [ gh + (1/2)(V^2)] / x^2

(NOTE : I trust you can do the required derivation of the above on your own as this is simple algebraic manipulation.)

and substituting appropriate values,

k = [2 (318)/95^2][(9.8 * 126(sqrt 3/2)) + (1/2)(11200)^2]

and performing the required arithmetic,

k = 4,420,011 N/m  = 4420.011 kN/m

(this is the required spring constant to satisfy the conditions of the problem)