A rifle fires a 2.11 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.37 10-2 m from its unstrained length. The pellet rises to a maximum height of 5.49 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

help me please =(

never mind!

I got it! =) thanks guys!

To determine the spring constant, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (spring and pellet) remains constant throughout the motion, assuming no external forces act on it.

At the maximum height, all the initial potential energy of the compressed spring is converted into the potential energy of the pellet at its highest point. The potential energy stored in a spring is given by the formula:

PE = (1/2)kx^2

where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the potential energy at the maximum height is equal to the potential energy stored in the compressed spring, so we have:

(1/2)kx^2 = mgh

where m is the mass of the pellet, g is the acceleration due to gravity, and h is the maximum height reached.

Now we can substitute the given values into the equation and solve for k:

m = 2.11 × 10^-2 kg
x = 8.37 × 10^-2 m
h = 5.49 m
g = 9.8 m/s^2

(1/2)k(8.37 × 10^-2)^2 = (2.11 × 10^-2)(9.8)(5.49)

Simplifying the equation:

(1/2)k(6.9969 × 10^-3) = 1.10263

k = (2 × 1.10263) / (6.9969 × 10^-3)
k = 314.7915 N/m

Therefore, the spring constant is approximately 314.7915 N/m.