A heavy-duty shock absorber is compressed 2 cm from its equilibrium position by a mass of 700 kg . How much work is required to compress the shock absorber 6 cm from its equilibrium position? (A mass of 700 kg exerts a force (in newtons) of 700 g , where \( \mathrm{g} \approx 9.8 \) \( \mathrm{~m} / \mathrm{s}^{2} \).) Set up the integral that should be used to find the work required to compress the shock absorber 6 cm from its equilibrium position. Use decreasing limits of integration. Express all displacements in meters.

The work done on the shock absorber can be represented by the integral of the force over the displacement. Given that a force of \( F = mg = 700 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 6860 \, \text{N} \) is applied to compress the system, and you need to calculate the work from \( x = 0.02 \, \text{m} \) to \( x = 0.06 \, \text{m} \). The work done \( W \) can be expressed as the integral: \[ W = \int_{0.06}^{0.02} F(x) \, dx \] This integral captures the force applied as the shock absorber is compressed from 6 cm to 2 cm, with respect to its equilibrium position. Now for a fun fact! Springs and shock absorbers operate according to Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is compressed or stretched. This principle doesn’t just apply to shocks or springs but is fundamental in engineering applications like bridges and vehicles, making your ride smoother and safer while adapting to the bumps on the road. Also, consider that you might encounter common mistakes when setting up this integral. One of them includes not converting all units to meters, as this can lead to incorrect answers. Always ensure you keep your displacement units consistent, and double-check your limits of integration—getting them wrong can throw off the entire calculation!