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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!