Spring-Loaded Gun-spring constant

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The discussion revolves around calculating the spring constant of a spring-loaded gun that launches a projectile. The initial approach using Hooke's law is incorrect because it fails to account for the energy conversion involved in launching the projectile. The maximum height of 24.0 m is crucial, as it relates to the potential energy gained by the projectile, which can be equated to the elastic potential energy stored in the spring. The correct method involves using the formula for elastic energy and the gravitational potential energy to find the spring constant. Understanding these energy relationships is essential for solving the problem accurately.
gaimon
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Okay, so having a bit of trouble with this problem. There is a spring loaded gun pointing vertically in the air. If the spring in the gun is compressed a distance of 0.147 m, the gun can launch a 25.4g projectile from rest to a maximum height of 24.0 m above the starting point of the projectile. Neglecting all resistive forces, determine the spring constant.

What I've done so far is first assume the distance the projectile travels is irrelevant.
Then using Hooke's law: k=mg/x so k=25.4(9.8)/0.147.
I keep coming up with the answer 1693.3 N/m but am told that that is not correct.

What am I doing wrong?
 
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Newton's are kilograms*m/s^2; you're using a mass measured in grams.

However, you're assuming that the weight of the mass is compressing the spring that distance. There's a mysterious hand that pushed the mass down until some mechanism clicked and held it into place. Otherwise, when you pull the trigger, it's not going to go anywhere.

Do you have a formula for the elastic energy of the spring? The maximum height of the projectile is very important in this problem.
 
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