Question

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?

Answer 1

x = 0.12 m

Step-by-step 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