Ask Question
22 April, 05:17

Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 5.40 10-2 kg and a charge of 5.10 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B

+4
Answers (1)
  1. 22 April, 06:46
    0
    0.247 J = 247 mJ

    Explanation:

    From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

    So, W = ΔK + ΔU = 1/2m (v₂² - v₁²) + q (V₂ - V₁) where m = mass of particle = 5.4 * 10⁻² kg, q = charge of particle = 5.10 * 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

    So, W = ΔK + ΔU = 1/2m (v₂² - v₁²) + q (V₂ - V₁)

    = 1/2 * 5.4 * 10⁻²kg * ((3m/s) ² - (2 m/s) ²) + 5.10 * 10⁻⁵ C (7850 - 5650)

    = 0.135 J + 0.11220 J

    = 0.2472 J

    ≅ 0.247 J = 247 mJ
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 5.40 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers