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6.7 Physical Applications (part 1)

long thin bar/wire

length is \( L \) m

A horizontal thin bar with endpoints labeled x=0 and x=L.

it's very thin so only the length matters

if the density is constant: \( \rho \) (rho)

then mass is \( m = \rho \cdot L \)

if density is not constant, say \( \rho = \rho(x) \)

A horizontal bar with a small vertical slice highlighted at position x, labeled with width dx.

this thin segment has mass = \( \rho(x) dx \)

to find the mass of entire thing we accumulate these by integration:

\[ m = \int_{0}^{L} \rho(x) dx \]

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example

wire length 6 m

density is twice the distance from the left end

mass = ?

A horizontal bar from x=0 to x=6 with a point x marked at a distance from the left end.

density here is \( \rho(x) = 2x \)

So mass of this wire is

\[ \int_{0}^{6} 2x dx = 36 \]

follow-up question: what is the mass of the right half of the wire?

the accumulation starts at \( x = 3 \) ends at \( x = 6 \)

\[ \int_{3}^{6} 2x dx = 27 \]
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Work and Variable Forces

Work is defined as:

\[ \text{work} = \text{force} \cdot \text{distance} \]

This assumes force is constant.

In many cases, the force is not constant. For example, a spring.

Force on a Spring: Hooke's Law

\[ F = kx \]
  • F: force
  • k: spring constant (stiffness)
  • x: deviation from the equilibrium position

At the equilibrium position, no force is involved. The length of the spring at equilibrium is its natural length.

Work with Non-Constant Force

The work in this case (non-constant force) is:

\[ W = \int_{a}^{b} \text{force} \cdot dx = \int_{a}^{b} kx \, dx \]
  • a: starting length measured with respect to natural length
  • b: ending length measured with respect to natural length
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Example: Spring Work Calculation

Example: A spring has a natural length of \( 1 \, \text{m} \).

A force of \( 40 \, \text{N} \) stretches and holds it at \( 0.1 \, \text{m} \) from its equilibrium.

Find:

  1. Work done in compressing from natural length to a length of \( 0.5 \, \text{m} \).
  2. Additional work done in compressing it by another \( 0.5 \, \text{m} \).

Step 1: Find the Spring Constant

The initial information tells us the spring constant:

\[ F = kx \]\[ 40 = k \cdot (0.1) \]\[ k = 400 \]

Note: A force of \( 40 \, \text{N} \) stretches it \( 0.1 \, \text{m} \) beyond equilibrium.

Step 2: Calculate Work to Compress

Work to compress from natural (\( x = 0 \)) to a length of \( 0.5 \, \text{m} \) (which corresponds to \( x = -0.5 \)):

\[ W = \int_{0}^{-0.5} 400x \, dx = \dots = 50 \, \text{N} \cdot \text{m} \text{ (Joules)} \]
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Work to Compress Another 0.5 m

The work required to compress the spring an additional 0.5 meters is calculated using the following integral:

\[W = \int_{-0.5}^{-1} 400x \, dx = \dots = 150 \text{ J}\]
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Work Done Against Gravity

If the change in height is small compared to the size of the Earth, then we can consider the acceleration due to gravity as constant:

\[g = 9.8 \text{ m/s}^2\]

\(\text{work} = \text{force} \cdot \text{distance}\)

Work in moving a mass of 45 kg from the ground to a height of 60 m is:

\[W = (45 \cdot g) \cdot 60 = 2700g\]

Note: Here \(g\) is approximately 9.8.

Variable Distance and Integration

Just like with force, if distance is not constant, then an integration is involved.

Example:

Work in winding up a chain or cable (distributed mass).