Answer to Question #184729 in Electricity and Magnetism for Mark John

Question

An electron is moving along an electric field. If the initial kinetic energy was greater than zero, describe the following parameters: Change in KE, Change in U, Change in V and work.

Answer 1

To be given in question

Kinetic energy $K_{i}>0$

To be asked in question

Change Kinetic energy change U and change V and Work

We know that we

kinetic energy KE $=\frac{1}{2}mv^2$

K.E =K_f-K_i $=\frac{1}{2}mv_{f}^2-\frac{1}{2}mv_{i}^2$

K.E=$=\frac{1}{2}mv_{f}^2-\frac{1}{2}mv_{i}^2>0$

$K.E=\frac{1}{2}mv_{f}^2>\frac{1}{2}mv_{i}^2$

K.E>0

Kinetic energy+potential energy =Total energy

Conservation system

Total energy=zero

Kinetic energy= - potential energy

K.E =-∆U=$-(mgh_{f}-mgh_{i})$

Put value

$h_{f}=h, h_{i}=0$ ( at ground)

$∆U=-(mgh-mg\times 0 )$

∆U= -mgh $=∆W=\frac{1}{2}mv^2 =K_{f}-K_{i}$

∆w=-mgh

Kinetic energy potential term

K.E=∆W=q∆v

K.E=q(v_f-v_i)=∆w

Work energy theorem

∆W=K.E=$\frac{1}{2}mv^2$

Work energy relation

$∆W=\frac{1}{2}mv^2 =K_{f}-K_{i}$