# A 1.5 kg box moves back and forth on a horizontal frictionless surface between two springs. The...

## Question:

A 1.5 kg box moves back and forth on a horizontal frictionless surface between two springs. The box is initially compressed against the stronger spring, K=32 N/cm, 4.0 cm and released from rest. The smaller box is K=16 N/cm.

A) By how much will the box compress the weaker spring?

B) What is the maximum speed the box will reach?

## Energy Conservation Principle:

The energy conservation principle consists of a physical law that states that if no external forces act on the system, the total energy of the system should conserve. Energy is not created nor destroyed, it only transforms. This means that a system may present energy transformations, but the overall total energy of the system will remain the same. The total energy of a system can be determined by the sum of the total kinetic and potential energy of the system. For elastic potential energy, the formula is:

{eq}U_e=\dfrac{1}{2}kx^2 {/eq}

#### Part a):

In order to find the compression in the weaker spring, we must use the energy conservation principle. The total energy on the strong spring will convert to the total energy on the weaker spring.

{eq}E_i=E_f\\ \dfrac{1}{2}k_1x_1^2=\dfrac{1}{2}k_2x_2^2\\ k_1x_1^2=k_2x_2^2\\ x_2=\sqrt{\dfrac{k_1x_1^2}{k_2}}\\ x_2=\sqrt{\dfrac{32\ N/cm\times 4^2\ cm^2}{16\ N/cm}}\\ x_2=5.66\ cm {/eq}

#### Part b):

The maximum speed will be reached at the equilibrium position. We use energy conservation mechanics to solve for the speed.

{eq}E_i=E_f\\ \dfrac{1}{2}k_1x_1^2=\dfrac{1}{2}mv^2\\ v=\sqrt{\dfrac{k_1x_1^2}{m}}\\ v=\sqrt{\dfrac{0.32\ N/m\times(0.04)^2\ m^2}{1.5\ kg}}\\ v=0.018\ m/s=1.8\ cm/s {/eq} 