Question

Question:(II) A 1200-kg car moving on a horizontal surface has speed\({v_i} = {\bf{2}}{\bf{.0 m/s}}\)when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Short answer

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

Step-by-step solution

Understanding the conservation of energy in the car and spring motion

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.

Identification of given data

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 \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Representation of the diagram of the car

The diagram can be shown as:

Here, k is the spring constant.

Analysis of conservation of energy

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 \(1.38 \times {10^5}{\rm{ N/m}}\).