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## Homework Statement

Two spheres are mounted on identical horizontal springs and rest on a frcionless table, as in the drawing. When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unrestrained. When each sphere has a charge of +1.60 microC, the spacing doubles. Assuming that the spheres have a negligible diameter,

**determine the spring constant of the springs.**

## Homework Equations

E=KE_t+KE_r+U_g+PE_sp+PE_e

Energy is equal to the sum of all energies, translational kinetic, rotational kinetic, gravitational potentional, potential spring, and potential electric.

Potential gravitational and Rotaional Kinetic are not applicable

KE_t=(1/2mv^2)

PE_sp=(1/2)kx^2

PE_e=[k(q_1)(q_2)]/r^2

## The Attempt at a Solution

E=KE_t+PE_sp+PE_e

Ef=[(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r_f^2]

Ei=0+[1/2kx_i^2]+[(k_e)(q^2)]/r_i^2]

Ef=Ei

[(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r-f^2]=[1/2kx_i^2]+[(k_e)(q^2)]/ri^2]

(1/2)[m_1f V_1f^2 +M_2f V_2f^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]= (1/2)[Kx_i^2]+[(2k_e)(q^2)]/r_i^2]

(m1=m2)

[mf Vf^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]=[(2k_e)(q^2)]/r_i^2]

Mv^2+x_f^2+(1/r_f^2)=(1/r_i^2)

So then...

(1/2) [m_1v_1f^2+m_2v_2f^2]+Kx_f^2+[(k_e)(q^2)]/r^2]=0

K=-[[k_e(q^2)/r^2]-[mv_f^2]]/x^2

K=-[(8.99e9)(2.56e6)]/(.01)]-m(1.73^2)

k=-29.929(x+7.689)

and I'm stuck...I don't know what I did wrong really...I don't know how to find the mass though...

Thanks!