To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 5.25 × 10^3 N m–1. What is the maximum compression of the spring ? Example 5.9 Consider Example 5.8 taking the coefficient of friction, µ, to be 0.5 and calculate the maximum compression of the spring.

Example 5.8 To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant $5.25\times 10^3 Nm^{-1}$ . What is the maximum compression of the spring ? [NCERT Class 11 Example 5.8]

Example 5.9 Consider Example 5.8 taking the coefficient of friction, µ, to be 0.5 and calculate the maximum compression of the spring. [NCERT Class 11 Example 5.9]

$5.25\times 10^3 Nm^{-1}$ corrected as $6.25\times 10^3 Nm^{-1}$

Without Friction: 

Concept: Law of Conservation of Energy, Without friction, all the KE of car will become PE of spring. Ideal case.  $x_m$ is the compression of the spring.

\[KE_{car}=PE_{spring}\]\[\frac{1}{2}mv^2=\frac{1}{2}k\hspace{1mm}x_m^2\]

\[\frac{1}{2}\hspace{1mm} 1000\hspace{1mm} kg. \left(5 \hspace{1mm}m/s\right)^2 = \frac{1}{2} \hspace{1mm}\left(6.25 \times10^3 \hspace{1mm}N/m\right)\hspace{1mm} x_m^2\]

\[x_m=2\hspace{1mm} m\]

With Friction: 

Concept: Law of Conservation of Energy, In reality, With friction, KE of car will go to PE of the spring and also work has to be done against friction.

\[KE_{car}=PE_{spring}+Work\hspace{1mm} done_{against\hspace{1mm} friction}\]

\[\frac{1}{2}mv^2=\frac{1}{2}k\hspace{1mm}x_m^2+F_{friction}. x_m\]

\[\frac{1}{2}mv^2=\frac{1}{2}k\hspace{1mm}x_m^2+\mu N . x_m\]

\[\frac{1}{2}mv^2=\frac{1}{2}k\hspace{1mm}x_m^2+\mu\hspace{1mm} mg . x_m\]

\[\frac{1}{2}\hspace{1mm} 1000\hspace{1mm} kg. \left(5 \hspace{1mm}m/s\right)^2 = \frac{1}{2} \hspace{1mm}\left(5.25 \times10^3 \hspace{1mm}N/m\right)\hspace{1mm} x_m^2+0.5\times 1000\hspace{1mm}kg\times 10\hspace{1mm}m/s^2 . x_m\]

\[25=6.25\hspace{1mm}x_m^2+10\hspace{1mm}x_m\]

\[1.25\hspace{1mm}x_m^2+2\hspace{1mm} x_m -5=0\]

Solve the Quadratic Equation to get,

\[x_m=\frac{-2\pm \sqrt{2^2-4\times 1.25\times \left(-5 \right) }}{2\times 1.25}\]

\[x_m=1.35\hspace{1mm} m\]

Compared to the previous case without friction, note that the maximum compression is reduced as some of the energy is spent in overcoming the friction. That's a reality check to see if our answer makes sense.