Question

A rifle fires a 2.10 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 9.10 x 10^-2 m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance and any other friction force, determine the spring constant.

Please be sure the answer is correct this means a lot to my grade this semester and I just can not seem to get it.

Answer 1

  Here the elastic potential energy of the spring is converted into potential energy.

0.5*K*x^2 = mgh

Thus K = 2mgh/x^2 = (2*2.10*10^-2*9.8*6.1)/(9.10*10^-2)^2

Thus K = 303.19N.m^-1

Answer 2

let spring constant be K.when the spring is compressed along with the pellet there is a potential enhergy associated with the spring given by PE=1/2Kx^2 where x is compression./

using values we get PE=4.14 X 10^-3K

now when the pellet is fired the spring force gives it a kinetic energy which is ultimately converted intoo gravitational potential energy GE of the pellet ..........GE is given by GE=MgH where M is pellet masss and H is height to which it rises..

equating GE and pe WE GET

MgH=PE

or 2.1 X 10^-2 X 9.81 X 6.1= 4.14 X 10^-3K

solving we get K=303.5N/m

Answer 3

spring constant = (2mgh/x^2) = (2*0.021*9.8*6.1/0.091^2) = 303.2 N/m