91Ó°ÊÓ

The car moving on the horizontal surface strikes the horizontal spring.According to the conservation of energy, the kinetic energy of the car is equal to the spring’s potential energy.The total energy is converted into the spring’s elastic potential energy.

The given data can be listed below as:

• The mass of the car is\(m = 1200{\rm{ kg}}\).
• The velocity of the car is\(v = 85{\rm{ km/h}}\left( {\frac{{1{\rm{ m/s}}}}{{3.6{\rm{ km/h}}}}} \right) = 23.61{\rm{ m/s}}\).
• The distance moved by the spring when the car strikes it is\(x = 2.2{\rm{ m}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The diagram can be shown as:

Here, k is the spring constant.

The kinetic energy of the car is equal to the spring potential energy. The energy can be expressed as:

\(\begin{array}{c}P.{E_i} + K.{E_i} = P.{E_f} + K.{E_f}\\0 + \frac{1}{2}m{v^2} = \frac{1}{2}k{x^2} + 0\\m{v^2} = k{x^2}\\k = \frac{{m{v^2}}}{{{x^2}}}\end{array}\)

Here,\(v\)is the velocity of the car. The initial gravitational potential energy\(P.{E_i}\)is equal to zero because the spring is not compressed, and the car and spring both lie at the same level.

\(K.{E_i}\)is the initial kinetic energy,\(P.{E_f}\)is the elastic potential energy of the spring, and\(K.{E_f}\)is the final kinetic energy, which is equal to zero after it strikes.

Substitute the values in the above expression.

\(\begin{array}{c}k = \frac{{1200{\rm{ kg}} \times {{\left( {23.61{\rm{ m/s}}} \right)}^2}}}{{{{\left( {2.2{\rm{ m}}} \right)}^2}}} \times \left( {\frac{{1{\rm{ N/m}}}}{{1{\rm{ kg/}}{{\rm{s}}^2}}}} \right)\\ = 1.38 \times {10^5}{\rm{ N/m}}\end{array}\)

Thus, the spring’s stiffness constant is \(1.38 \times {10^5}{\rm{ N/m}}\).