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 rough road having $\mu$ to be $0.5$ and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{3} \;N m ^{-1} .$ What is the maximum compression of the spring in $m$?
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 rough road having $\mu$ to be $0.5$ and colliding with a horizontally mounted spring of spring constant $6.25 \times 10^{3} \;N m ^{-1} .$ What is the maximum compression of the spring in $m$?
Answer In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Figure
We invoke the work-energy theorem, rather than the conservation of mechanical energy. The change in kinetic energy is
$\Delta K=K_{f}-K_{t}=0-\frac{1}{2} m v^{2}$
The work done by the net force is
$W=-\frac{1}{2} k x_{m}^{2}-\mu m g x_{m}$
Equating we have
$\frac{1}{2} m v^{2}=\frac{1}{2} k x_{m}^{2}+\mu m g x_{m}$
Now $\mu m g=0.5 \times 10^{3} \times 10=5 \times 10^{3} N$ (taking
$\left.g=10.0 m s ^{-2}\right) .$ After rearranging the above equation we obtain the following quadratic equation in the unknown $x_{m}$
$k x_{m}^{2}+2 \mu m g x_{m}-m v^{2}=0$
$x_{m}=\frac{-\mu m g+\left[\mu^{2} m^{2} g^{2}+m k v^{2}\right]^{1 / 2}}{k}$
where we take the positive square root since $x_{m}$ is positive. Putting in numerical values we obtain
$x_{m}=1.35 m$
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- [JEE MAIN 2022]