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 $6.25 \times 10^{3} \;N m ^{-1} .$ 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 smooth road 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 At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring. The kinetic energy of the moving car $1 \,s$
$K=\frac{1}{2} m v^{2}$
$=\frac{1}{2} \times 10^{3} \times 5 \times 5$
$K=1.25 \times 10^{4} J$
where we have converted $18 km h ^{-1}$ to $5 m s ^{-1}$ [It is useful to remember that $36 km h ^{-1}=10 m s ^{-1} $ ]
At maximum compression $x_{m^{\prime}}$, the potential energy $V$ of the spring is equal to the kinetic energy $K$ of the moving car from the principle of conservation of mechanical energy.
$V=\frac{1}{2} k x_{m}^{2}$
$=1.25 \times 10^{4} J$
We obtain
$x_{m}=2.00\; m$
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.
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