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

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

Step-by-step 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