Many cars have “\(5\;{\rm{mi/h}}\;\left( {8\;{\rm{km/h}}} \right)\)bumpers” that are designed to compress and rebound elastically without any physical damage at speeds below \(8\;{\rm{km/h}}\). If the material of the bumpers permanently deforms after a compression of 1.5 cm, but remains like an elastic spring up to that point, what must be the effective spring constant of the bumper material, assuming the car has a mass of 1050 kg and is tested by ramming into a solid wall?

According to the work-energy theorem, the total work done on a particle is equal to the change in its kinetic energy. Mathematically,

\({W_{{\rm{net}}}} = \Delta KE\)

It is also known as the principle of work and energy.

The speed of the car is \(v = {\rm{8}}\;{\rm{km/h}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}} \times \frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}} = 2.22\;{\rm{m/s}}\).

The compression in the spring is \(x = 1.5\;{\rm{cm}} = 0.015\;{\rm{m}}\).

The mass of the car is \(m = 1050\;{\rm{kg}}\).

The total work done on the car is equal to the elastic potential energy stored in the spring (bumper) and the initial kinetic energy of the car.

The work done is negative because the impact on the car is in the opposite direction of the displacement of the car. Mathematically,

\(\begin{aligned}{W_{{\rm{spring}}}} &= \Delta KE\\ - \Delta PE &= \Delta KE\\ - \left( {\frac{1}{2}k{x^2} - 0} \right) &= \left( {\frac{1}{2}m{{\left( 0 \right)}^2} - \frac{1}{2}m{v^2}} \right)\\k{x^2} &= m{v^2}\end{aligned}\) … (i)

Rearranging equation (i) gives the required expression for the spring constant.

\(\begin{aligned}k &= \frac{{m{v^2}}}{{{x^2}}}\\ &= \frac{{\left( {1050\;{\rm{kg}}} \right){{\left( {2.22\;{\rm{m/s}}} \right)}^2}}}{{{{\left( {0.015\;{\rm{m}}} \right)}^2}}}\\ &= 2.3 \times {10^7}\;{\rm{N/m}}\end{aligned}\)

Thus, the effective spring constant is \(2.3 \times {10^7}\;{\rm{N/m}}\).