In a TV picture tube, electrons are emitted from the cathode with negligible speed and attain a velocity of 5 × 10^ 6 m/s after travelling one centimeter. Assuming straight line motion, determine the constant force

Correct answer: C. 1.1 × 10^ -15 N Given initial velocity u = 0 , final velocity v = 5 × 10^6 m/s , distance s = 1 cm = 10^ -2 m , and mass m = 9.1 × 10^ -31 kg . Using the equation of motion v^2 = u^2 + 2as : a = (v^2 - u^2)/(2s) a = (5 × 10^6)^2 - 0 2 × 10^ -2 a = 25 × 10^ 12 2 × 10^ -2 = 1.25 × 10^ 15 m/s^2 The constant force acting on the electron is given by Newton's second law, F = ma : F = (9.1 × 10^ -31 ) × (1.25 × 10^ 15 ) F = 11.375 × 10^ -16 N F 1.1 × 10^ -15 N

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Newton's Laws of Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Newton's Laws of Motion

In a TV picture tube, electrons are emitted from the cathode with...

In a TV picture tube, electrons are emitted from the cathode with negligible speed and attain a velocity of

5 \times 10^{6} \text{ m/s}

after travelling one centimeter. Assuming straight line motion, determine the constant force acting on the electron. The mass of the electron is

9.1 \times 10^{-31} \text{ kg}

Options

A $1.1 \times 10^{-17} \text{ N}$
B $2.3 \times 10^{-15} \text{ N}$
C $1.1 \times 10^{-15} \text{ N}$
D $4.6 \times 10^{-22} \text{ N}$

Correct answer

C. $1.1 \times 10^{-15} \text{ N}$

Step-by-step solution

Given initial velocity

u = 0

, final velocity

v = 5 \times 10^6 \text{ m/s}

, distance

s = 1 \text{ cm} = 10^{-2} \text{ m}

, and mass

m = 9.1 \times 10^{-31} \text{ kg}

.

Using the equation of motion

v^2 = u^2 + 2as

a = \dfrac{v^2 - u^2}{2s}

a = \dfrac{(5 \times 10^6)^2 - 0}{2 \times 10^{-2}}

a = \dfrac{25 \times 10^{12}}{2 \times 10^{-2}} = 1.25 \times 10^{15} \text{ m/s}^2

The constant force acting on the electron is given by Newton's second law,

F = ma

F = (9.1 \times 10^{-31}) \times (1.25 \times 10^{15})

F = 11.375 \times 10^{-16} \text{ N}

F \approx 1.1 \times 10^{-15} \text{ N}

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