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

i think they are all correct

explanation:

they seem good

need to see circuit to get time constant