Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be [tex]K_A[/tex] = 1.9000000000000001 J
and the final kinetic energy from point B be [tex]K_B[/tex] = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
[tex]-Q \times ( V_B - V_A) = (K_B - K_A)[/tex]
[tex]-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]15 = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]K_B = 15+ 1.9000000000000001 \ J[/tex]
[tex]\mathbf{K_B =1 6.9000000000000001 \ J}[/tex]
