# Box and Spring

A 2.5 box that is sliding on a frictionless surface with a speed of 13 approaches a horizontal spring. The spring has a spring constant of 2100 N/m .

(a) If one end of the spring is fixed and the other end changes its position, how far will the spring be compressed in stopping the box?

(b)How far will the spring be compressed when the box’s speed is reduced to half of its initial speed?

One must assume that the box has mass m=2.5 kg and speed v=13 m/s.

a) The total energy of the system box + spring is the same initial and final.

Initial the energy is kinetic (box energy) $Ec = m*v^2/2$

Final energy is potential (compressed string energy) $Ep =k*x^2/2 $where $x$ is the compression distance of the spring.

$mv^2/2 =k*x^2/2$

$x = v*\sqrt{m/k} =13*\sqrt{2.5/2100} =0.448 m$

b) again the total energy is the same . initial speed this time $v = 13/2 =6.5 m/s$

$mv^2/2 =k*x^2/2$

$x = v*\sqrt{m/k} =6.5*\sqrt{2.5/2100} =0.224 m$