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

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 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

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Mít Mít 3 years 2021-08-22T01:47:52+00:00 1 Answers 103 views 0

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  1. Helga
    0
    2021-08-22T01:49:24+00:00
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    Answer:

    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

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