Question

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 1

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 2

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