A rifle fires a 2.90 x 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 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 7.23 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answer: 587.14 N/m

Explanation:

E (f) = E (0) such that

0.5mv (f) ² + 0.5Iw (f) ² + mgh (f) + 0.5ky (f) ² =

0.5mv (0) + 0.5Iw (0) ² + mgh (0) + 0.5ky (0) ²

since the pellets does not rotate, then the angular speeds are zero, so, w (f) and w (0) = 0

Since the pellet is at rest, and sits on the Spring, the translational speed, v (0) and v (f) are zero too.

Since the Spring is not strained when it reaches maximum height, y (f) = 0, so that

mgh (f) = mgh (0) + 0.5ky (0) ²

[mgh (f) - mgh (0) ] / y² = 0.5k

mg[h (f) - h (0) ] / y² = 0.5k

(2.9*10^-2 * 9.8 * 7.23) / (8.37*10^-2) ² = 0.5k

2.055 / 7*10^-3 = 0.5k

0.5k = 293.571

k = 293.571/0.5

k = 587.14 N/m

Answer: 586.60N/m

Explanation:

In this scenario, the elastic potential energy of the spring is converted into potential energy.

0.5*K*x^2 = mgh

Thus K = 2mgh/x^2

= (2*2.90*10^-2*9.8*7.23) / (8.37*10^-2) ^2

=586.599

Therefore K = 586.60N/m