Chapter 9: Q. 46 (page 229)

A particle of mass m moving along the $x -$ axis has velocity $v_{x} = v_{0} \sin (πx / 2 L)$ . How much work is done on the particle as it moves (a) from $x = 0$ to $x = L$ and
(b) from $x = 0$ to $x = 2 L$ ?

Short Answer

Expert verified

a). $\frac{m {v_{0}}^{2}}{4}$

b). $\frac{- m {v_{0}}^{2}}{4}$

Step by step solution

Step 1. Given Information

Mass of particle $= m$

Velocity of the particle $v_{x} = v_{o} \sin (πx / 2 L)$

Step 2. Part a). Work done on the particle

When it moves from $x = 0$ to $x = L$ . For that the we will proceed as

$W = \int_{a}^{b} F d x W = \int_{a}^{b} m a d x a_{x} = v_{x} \frac{d v_{x}}{d x} \frac{d v_{x}}{d x} = v_{0} \frac{π}{2 L} \cos \frac{πx}{2 L}$

Now,

$W = \int_{a}^{b} m a d x = \int_{0}^{L} m {v_{0}}^{2} \frac{π}{2 L} \sin \frac{πx}{2 L} \cos \frac{πx}{2 L} d x = m {v_{0}}^{2} \frac{π}{4 L} \int_{0}^{L} 2 \sin \frac{πx}{2 L} \cos \frac{πx}{2 L} d x = m {v_{0}}^{2} \frac{π}{4 L} \int_{0}^{L} \sin \frac{πx}{L} d x$

Using identity: $\sin 2 θ = 2 \sin θ \cos θ$

$= - m {v_{0}}^{2} \frac{1}{4} (\cos \frac{πL}{L} - \cos 0) = \frac{m {v_{0}}^{2}}{4}$

Step 3. Part b). Work done on the particle

We need to determine work done on the particle when it form $x = 0$ to $x = 2 L$

Now,

$W = \int_{a}^{b} m a d x = \int_{0}^{2 L} m {v_{0}}^{2} \frac{π}{2 L} \sin \frac{πx}{2 L} \cos \frac{πx}{2 L} d x = m {v_{0}}^{2} \frac{π}{4 L} \int_{0}^{2 L} 2 \sin \frac{πx}{2 L} \cos \frac{πx}{2 L} d x = m {v_{0}}^{2} \frac{π}{4 L} \int_{0}^{L} \sin \frac{πx}{L} d x = {- m {v_{0}}^{2} \cos \frac{πx}{L}}^{2 L} = - \frac{m {v_{0}}^{2}}{4} (\cos \frac{2 πL}{L} - \cos 0) = - \frac{m {v_{0}}^{2}}{4}$

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A particle of mass m moving along the $x -$ axis has velocity $v_{x} = v_{0} \sin (πx / 2 L)$ . How much work is done on the particle as it moves (a) from $x = 0$ to $x = L$ and
(b) from $x = 0$ to $x = 2 L$ ?

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Part a). Work done on the particle

Step 3. Part b). Work done on the particle

One App. One Place for Learning.

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A particle of mass m moving along the x-axis has velocity vx=v0sin(πx/2L). How much work is done on the particle as it moves (a) from x=0to x=Land (b) from x=0tox=2L?

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Part a). Work done on the particle

Step 3. Part b). Work done on the particle

One App. One Place for Learning.

Most popular questions from this chapter

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A particle of mass m moving along the $x -$ axis has velocity $v_{x} = v_{0} \sin (πx / 2 L)$ . How much work is done on the particle as it moves (a) from $x = 0$ to $x = L$ and
(b) from $x = 0$ to $x = 2 L$ ?