Question

A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s2 (b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval? m (d) What is its kinetic energy at the end of this interval? J

Answer 1

Answer:

a)  a = 6.31 10¹⁰ m / s² , b) t = 2.06 10⁻⁵ s , c)  x = 13.39 m , d)  ΔK = 1.41 10⁻¹⁵ J

Explanation:

a) Since they indicate that the speeds are non-relativistic, we can use the kinematics relations and Newton's second law

                  F = m a

                 

The force in electrical is

                 F = qE

                qE = m a

                a = qE / m

we calculate

                a = 1.6 10⁻¹⁹ 660 / 1.673 10⁻²⁷

                a = 6.31 10¹⁰ m / s²

b) Let's use the one-dimensional kinematics relation

                 v = v₀ + a t

as part of rest its initial velocity is zero

                 v = a t

                 t = v / a

                 t = 1.30 10⁶ / 6.31 10¹⁰

                 t = 2.06 10⁻⁵ s

c) We use the kinematics displacement equation

                 x = v₀ t + ½ a t²

initial velocity is zero

                x = ½ a t²

                x = ½ 6.31 10¹⁰ (2.06 10⁻⁵)²

                x = 1,339 10¹ m

                x = 13.39 m

d) the kinetic energy is

            ΔK = Kf -K₀

            ΔK = ½ m v² - 0

            ΔK = ½ 1.673 10⁻²⁷ (1.30 10⁶) 2

            ΔK = 1.41 10⁻¹⁵ J