A​ heavy-duty shock absorber is compressed 2 cm from its equilibrium position by a mass of 500nbspkg. How much work is required to compress the shock absorber 4 cm from its equilibrium​ position? (A mass of 500 kg exerts a force​ (in newtons) of 500​g, where galmost equals9.8 m divided by s squared​.)
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Part 1
Set up the integral that should be used to find the work required to compress the shock absorber 4 cm from its equilibrium position. Use decreasing limits of integration. Express all displacements in meters.
Integral from nothing to nothing left parenthesis nothing right parenthesis dy

To solve this problem, we need to calculate the work required to compress the shock absorber further from its equilibrium position. We know that a mass of 500 kg compresses the shock absorber by 2 cm, and the work done is related to the force applied and the displacement.

### Step 1: Determine the Force Constant (k)

The force \( F \) exerted by the mass is given by:

\[
F = 500 \times g = 500 \times 9.8 = 4900 \, \text{N}
\]

This force compresses the shock absorber by 2 cm (or 0.02 m). Since the shock absorber behaves like a spring, we can use Hooke's Law to determine the spring constant \( k \):

\[
F = kx
\]

where \( x \) is the displacement from the equilibrium position in meters. Substituting the values:

\[
4900 = k \times 0.02
\]

Solving for \( k \):

\[
k = \frac{4900}{0.02} = 245000 \, \text{N/m}
\]

### Step 2: Set Up the Integral for Work

The work required to compress a spring from position \( x_1 \) to \( x_2 \) is given by the integral:

\[
W = \int_{x_1}^{x_2} kx \, dx
\]

where:
- \( x_1 \) is the initial compression (0 m).
- \( x_2 \) is the final compression (0.04 m).

The limits of integration should be set from \( x = 0.04 \, \text{m} \) to \( x = 0 \, \text{m} \), because we are compressing from the equilibrium (0) to the compressed position (0.04 m). So, the integral becomes:

\[
W = \int_{0.04}^{0} 245000x \, dx
\]

or equivalently (using decreasing limits):

\[
W = -\int_{0}^{0.04} 245000x \, dx
\]

### Step 3: Evaluate the Integral

Now, evaluate the integral:

\[
W = -\int_{0}^{0.04} 245000x \, dx = - \left[ \frac{245000x^2}{2} \right]_{0}^{0.04}
\]

\[
W = - \left[ \frac{245000 \times (0.04)^2}{2} - \frac{245000 \times (0)^2}{2} \right]
\]

\[
W = - \left[ \frac{245000 \times 0.0016}{2} \right]
\]

\[
W = - \left[ \frac{392}{2} \right] = -196 \, \text{J}
\]

However, since work is scalar and typically considered in positive terms (as energy required), we take the absolute value:

\[
W = 196 \, \text{J}
\]

Therefore, the work required to compress the shock absorber 4 cm from its equilibrium position is **196 J**.

Would you like to go over any part in more detail or have additional questions? Here are some related questions you might find interesting:

1. How is the spring constant related to the stiffness of a spring?
2. How does the amount of work change if the compression is doubled again?
3. What would the work be if the spring constant was different?
4. How does Hooke's Law apply to real-world shock absorbers?
5. What are the energy transformations involved in this compression process?

**Tip:** Remember, when dealing with spring or shock absorber problems, always ensure your displacement units match the units used for the spring constant to keep calculations accurate.

A heavy-duty shock absorber is compressed 2 cm from its equilibrium position by a mass of 500 kg. How much work is required to compress the shock absorber 4 cm from its equilibrium​ position? (A mass of 500 kg exerts a force​ (in newtons) of 500​g, where g almost equals 9.8 m/s².)

Learn how to calculate the work required to compress a shock absorber using Hooke's Law and integration. The problem involves determining the spring constant and setting up an integral to find the work done. Suitable for advanced high school or undergraduate physics students.

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