Ch 37: Special Relativity

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 37: Special Relativity

Problem 29b

Chapter 37, Problem 29b

A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

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Start by recalling the relativistic expression for force: $ F = \frac{dp}{dt} $, where $ p $ is the relativistic momentum given by $ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $. Here, $ m $ is the rest mass of the particle, $ v $ is its velocity, and $ c $ is the speed of light.

For a given acceleration $ a $, the force can be expressed as $ F = \frac{d}{dt} \left( \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \right) $. However, for simplicity, we can use the effective mass concept: $ F = ma_{\text{eff}} $, where $ a_{\text{eff}} = \frac{a}{(1 - \frac{v^2}{c^2})^{3/2}} $.

At rest ($ v = 0 $), the force required to produce the acceleration $ a $ is $ F_0 = ma $, since $ (1 - \frac{v^2}{c^2})^{3/2} = 1 $.

When the particle is moving at a velocity $ v $, the force required becomes $ F_v = \frac{ma}{(1 - \frac{v^2}{c^2})^{3/2}} $. The problem states that $ F_v = 2F_0 $, so substitute $ F_0 = ma $ into this equation: $ \frac{ma}{(1 - \frac{v^2}{c^2})^{3/2}} = 2ma $.

Simplify the equation to solve for $ v $: $ \frac{1}{(1 - \frac{v^2}{c^2})^{3/2}} = 2 $. Raise both sides to the power of $ -2/3 $ to isolate $ 1 - \frac{v^2}{c^2} $, and then solve for $ v $ in terms of $ c $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). This principle is fundamental in understanding how forces affect the motion of particles, particularly in determining the relationship between force, mass, and acceleration in both stationary and moving states.

Relativistic Effects

As an object's speed approaches the speed of light, relativistic effects become significant, altering the relationship between mass, force, and acceleration. Specifically, the effective mass of an object increases with speed, which means that more force is required to achieve the same acceleration as the object moves faster, a concept crucial for understanding the problem at hand.

Acceleration and Velocity Relationship

Acceleration is defined as the rate of change of velocity over time. When a force is applied to a particle in motion, the resulting acceleration depends not only on the force applied but also on the particle's current velocity. This relationship is essential for determining how the required force changes when the particle is already in motion compared to when it is at rest.

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