A stationary block has a charge of +6.0×10−8 C. A 0.80-kg cart with a charge of +4.0×10−8 C is initially at rest and separated from the block by 0.4 m. The cart is released and moves along a frictionless surface to a distance of 1.0 m from the block. Consider the objects as point charges.

A) Determine the electric potential energy of the initial state.
B)Determine the electric potential energy of the final state.
C)Determine the change in electric potential energy.
D)Determine the final speed of the cart. Assume that the change in electric potential energy is fully transformed into kinetic energy of the cart.

(a) U = 5.40 × 10^-7 J

(b) U = 2.16 × 10^-7 J

(c) ∆ U = 3.24 × 10^-7 J

(d) v = 9.0 × 10^-4 m/s

Explanation:

Electric Potential Energy, U = kq1q2/d

K = 9 × 10^9 Nm²/C², q1 = 6.0 × 10^-8 C, q2 = 4.0 × 10^-8 C, d = 0.4 m

(a) U = (9 × 10^9 × 6.0 × 10^-8 × 4.0 × 10^-8) / 0.4

U = 5.40 × 10^-7 J

(b) At d = 1.0 m,

U = (9 × 10^9 × 6.0 × 10^-8 × 4.0 × 10^-8) / 1.0

U = 2.16 × 10^-7 J

(c) ∆ U = (5.40 × 10^-7) - (2.16 × 10^-7)

∆ U = 3.24 × 10^-7 J

(d) ∆ U = ½mv²; where m = 0.80 Kg, ∆ U = 3.24 × 10^-7 J

v = √(2 × ∆ U / m)

v = √(2 × 3.24 × 10^-7 / 0.8)

v = 9.0 × 10^-4 m/s

answer

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explanation:

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