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:

$W = \int_{x_{i}}^{x_{f}} F_{x} d x$

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" $W = \int_{0}^{L} F_{x} d x W = \int_{0}^{L} F_{0} e^{- x / L} d x W = F_{0} {(- L e^{- x / L})}^{L}_{0} W = F_{0} L (1 - \frac{1}{e})$

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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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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

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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 $x = L$ ?