# A 1620 kg car rolling on a horizontal surface has a speed of 70 km/hr when it strikes a...

## Question:

A 1620 kg car rolling on a horizontal surface has a speed of 70 km/hr when it strikes a horizontal coiled spring and is brought to rest at a distance of 5.5 m. What is the spring constant (in N/m) of the spring? (Ignore non-conservative forces such as friction.)

## Spring Constant:

When a spring is compressed from its unstretched position by the application of an external force, the energy stored in the spring is called the elastic potential energy. The elastic potential energy stored in the spring is directly proportional to the stiffness of the spring or the spring constant. The spring constant is expressed in {eq}\rm N/m {/eq}.

## Answer and Explanation: 1

We are given the following data:

• Mass of the car, {eq}m=1620\ \text{kg} {/eq}
• Speed of rolling, {eq}V=70\ \text{km/h} {/eq}
• length of compression of the spring, {eq}x=5.5\ \text{m} {/eq}

As the car is rolling on a horizontal surface at a speed ({eq}(V) {/eq}, the kinetic energy in the car due to its motion can be calculated by using the following equation:

{eq}E=\dfrac{1}{2}mV^{2} {/eq}

Where

• m is the mass of the car
• V is the speed of the car

Substituting values in the above equation, we have:

{eq}\begin{align} E&=\dfrac{1}{2}mV^{2}\\[0.3 cm] &=\dfrac{1}{2}\times1620\ \text{kg}\times70\ \text{km/h}&\left [\text{1 km=1000 m, 1 h=3600 s} \right ]\\[0.3 cm] &=\dfrac{1}{2}\times1620\ \dfrac{\text{N}}{\text{m/s}^{2}}\times\left (\dfrac{70\times1000\ \text{m}}{1\times3600\ \text{s}} \right )^{2}\\[0.3 cm] &=306250\ \rm{N\cdot m} \end{align} {/eq}

Neglecting non-conservative forces, the whole kinetic energy is transformed into the elastic potential energy of the spring as the car stop. The elastic potential energy of the spring is expressed by the following equation:

{eq}E'=\dfrac{1}{2}kx^{2} {/eq}

Where

• k is the spring constant
• x is the length of compression of the spring

Substituting values in the above equation, we have:

{eq}\begin{align} E'&=\dfrac{1}{2}kx^{2}\\[0.3 cm] 306250\ \rm{N\cdot m}&=\dfrac{1}{2}\times k\times(5.5\ \text{m})^{2}\\[0.3 cm] k&=\frac{306250\ \rm{N\cdot m}\times2}{(5.5\ \text{m})^{2}}\\[0.3 cm] &\approx\boxed{20\times10^3\ \text{N/m}} \end{align} {/eq}

#### Learn more about this topic:

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.1K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.