Q.Consider Example 5.8 taking the coefficient of friction, $\mu$, to be $0.5$ and calculate the maximum compression of the spring.

Concept: Conservation of Mechanical Energy (with work done against friction)

Data reused from the previous spring-collision example (Q. "Examples 5.9"): a car of mass $m = 1000\,\text{kg}$, moving at $v = 18.0\,\text{km/h} = 5.00\,\text{m/s}$, collides with a horizontally mounted spring of spring constant $k = 5.25 \times 10^{3}\,\text{N m}^{-1}$. (Check: frictionless, this gives the compression $x_0 = v\sqrt{m/k} = 5.00\sqrt{1000/5250} \approx 2.18\,\text{m}$ — matching the previous example, confirming these are the right values to carry forward.)

Step 1 — Energy balance with friction

Now the road has friction, with coefficient $\mu = 0.5$, acting over the compression distance $x$. The car's initial kinetic energy is used up by (a) friction, which does negative work $-\mu m g x$, and (b) the spring, which stores $\frac12 k x^2$ at maximum compression (where the car is momentarily at rest):

Step 2 — Substitute values (using $g = 9.8\,\text{m/s}^2$):

Step 3 — Solve the quadratic …

Recovering the referenced example's data

This question asks us to redo the previous spring-collision example with friction added, so we must first pin down what that example gives: a car of mass $m = 1000\,\text{kg}$ moving at speed $v = 18.0\,\text{km/h} = 5.00\,\text{m/s}$ on a horizontal, smooth road, colliding with a spring of spring constant $k = 5.25\times10^3\,\text{N/m}$. As a sanity check, the frictionless answer to that example is

which matches the previous example's result — confirming these are the correct values to carry forward.

Setting up the energy equation with friction

Now the road has friction, with coefficient $\mu = 0.5$. As the car slides forward and compresses the spring, two things remove its kinetic energy: the spring (which stores it as elastic potential energy) and friction (which removes it permanently as heat). At the point of maximum compression $x$, the car is momentarily at rest, so all its initial kinetic energy has gone into these two channels:

The left side is the car's initial kinetic energy. The right side is the elastic potential energy stored in the spring plus the energy dissipated by friction over the distance $x$ that the car travels while compressing the spring.

Step-by-step solution

1. Write down the known quantities

• $m = 1000\,\text{kg}$
• $v = 18.0\,\text{km/h} = 5.00\,\text{m/s}$
• $k = 5.25 \times 10^3\,\text{N/m}$
• $\mu = 0.5$
• $g = 9.8\,\text{m/s}^2$

2. Substitute into the energy equation

3. Rearrange into standard quadratic form …

Concept: Energy Conservation with Non-Conservative Friction (Work-Energy + Spring PE)

Step 1: Reuse the previous spring-collision example's data, now with friction added.

$m = 1000\,\text{kg}$, $v = 5.00\,\text{m/s}$ (from $18.0\,\text{km/h}$), $k = 5.25\times10^3\,\text{N/m}$, $\mu = 0.5$, $g=9.8\,\text{m/s}^2$.

Step 2: Write the energy balance at maximum compression $x$ (car momentarily at rest).

Kinetic energy is split between the spring's stored PE and the energy dissipated by friction over distance $x$:

Step 3: Substitute the numbers and simplify. …