A 0.299 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