A0.299 kg mass slides on a frictionless floor with a speed of 1.44 m/s. the mass strikes and compresses a spring with a force constant of 44.9 n/m. how far does the mass travel after contacting the spring before it comes to rest?

x = 0.12 m

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

             x^2 = [(m)(v^2)]/k

                 x =  \sqrt{ [(m)(v^2)]/k}

                 x = \sqrt{ [(0.299)((1.44)^2)]/(44.9)}

                 x = 0.12 m

Therefore, the mass will travel 0.12 m and come to rest.

                    = 0.491 m/s

igneous rocks also form where plates converge. the subducting plate melts as it sinks into the crust of the earth, and the melt rises into the overriding plate forming volcanoes. metamorphic rocks are formed mainly in the lithosphere, wherever there is high pressure and high temperature. -woods

snell's law

refractve index = sin incidence / sin refraction