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A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100 J and its kinetic energy is 10 J. What is the particle's potential energy when its velocity reaches 4 m/s?

Respuesta :

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

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