Q.52 The force acting on a particle i... [FREE SOLUTION]

Chapter 9: Q.52 (page 229)

The force acting on a particle is $F x = F_{0} e^{- x / L}$ . How much work does this force do as the particle moves along the x-axis from $x = 0$ to $x = L$ ?

Short Answer

Expert verified

The work done on the particle as it moves from $x = 0 t o x = L$ is $F_{0} L (1 - \frac{1}{e}) .$

Step by step solution

Content Introduction

The work done by the force $F_{x}$ on the particle if the movement of particle is in same direction as of the force can be expressed as:

$W = {鈭�}_{x_{i}}^{x_{f}} F_{x} d x$

Here, $F_{x}$ is the force and $d x$ is the small displacement in the direction of the force.

Content Explanation

The work done on the particle as it moves from $x = 0 t o x = L$ is shown as follow:

role="math" localid="1647794367778" $W = {鈭�}_{0}^{L} F_{x} d x W = {鈭�}_{0}^{L} F_{0} e^{- x / L} d x W = F_{0} {(- L e^{- x / L})}^{L}_{0} W = F_{0} L (1 - \frac{1}{e})$

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91影视

The force acting on a particle is $F x = F_{0} e^{- x / L}$ . How much work does this force do as the particle moves along the x-axis from $x = 0$ to $x = L$ ?

Short Answer

Step by step solution

Content Introduction

Content Explanation

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91影视

The force acting on a particle is Fx=F0e-x/L . How much work does this force do as the particle moves along the x-axis from x=0 to x=L?

Short Answer

Step by step solution

Content Introduction

Content Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Space Physics

Dynamics

Linear Momentum

Measurements

Turning Points in Physics

Astrophysics

Study anywhere. Anytime. Across all devices.

The force acting on a particle is $F x = F_{0} e^{- x / L}$ . How much work does this force do as the particle moves along the x-axis from $x = 0$ to $x = L$ ?