Question

A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the proton at the end of the motion

Answer 1

The kinetic energy is
`K_f = 1.424 *10^(-15) \ J`

Step-by-step explanation:

From the question we are told that

The initial speed of the proton is
`v_p = 3.0 * 10^(5) \ m/s`

The distance covered is
`d = 3.5 \ m`

The magnitude of the electric field is
`E = 120\ N/C`

Generally the mass of a proton is
`m = 1.67 *10^(27) \ kg`

and the charge on a proton is
`q= 1.60*10^(-19) \ C`

Now according to work energy theorem,

`Work \ Done(W) = Kinetic \ Energy \ Change`

=>
`W = K_f -K_i`

=>
`K_f = W +K_i`

Where
`K_f` is final kinetic energy and
`K_i` is initial kinetic energy which is mathematically represented as

`K_i = (1)/(2) * m * v_p^2`

Now the net workdone(W) is mathematically represented as

`W = F * d = q* E* d`

So

`K_f = q* E * d + (1)/(2) * m * v_p^2`

substituting values

`K_f = 1.60*10^(-19)* 120 * 3.5 + (1)/(2) * 1.67*10^(-27) *(3.0*10^(5))^2`

`K_f = 1.424 *10^(-15) \ J`