Question

A rifle fires a 2.01 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 9.45 10-2 m from its unstrained length. The pellet rises to a maximum height of 6.03 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answer 1

Answer:

The value of spring constant is 266.01 $\frac{N}{m}$

Explanation:

Given:

Mass of pellet $m = 2.01 \times 10^{-2}$ kg

Height difference of pellet rise $h_{f} - h_{o} = 6.03$ m

Spring compression $x = 9.45 \times 10^{-2}$ m

From energy conservation law,

Spring potential energy is stored into potential energy,

  $mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}$

Where $k =$ spring constant, $g = 9.8 \frac{m}{s^{2} }$

  $k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }$

  $k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }$

  $k = 266.01$ $\frac{N}{m}$

Therefore, the value of spring constant is 266.01 $\frac{N}{m}$