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To stimulate car accidents auto manufacturers study the collision of moving cars with mounted...

Question:

To stimulate car accidents auto manufacturers study the collision 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 5 m/s on a smooth road and colliding with a horizontally mounted spring of spring const -6250. What is the maximum compression of the spring 1.when there is no friction and 2. when μ=0.5?

Spring Constant:

It is the effect of the object's material that opposes any change in the object's equilibrium position. This effect varies with the material, applied force and change in the equilibrium position of the object.

Answer and Explanation: 1


Given Data

  • The mass of the car is: {eq}m = 1000\;{\rm{kg}} {/eq}.
  • The speed of the car is: {eq}v = 5\;{\rm{m}}/{\rm{s}} {/eq}.
  • The spring constant of the spring is: {eq}k = 6250\;{\rm{N}}/{\rm{m}} {/eq}.
  • The coefficient of friction between car and road is: {eq}\mu = 0.5 {/eq} .


Apply the work energy theorem to calculate the maximum compression of the spring.

{eq}\begin{align*} \dfrac{1}{2}m{v^2} &= \dfrac{1}{2}k{x^2} + \mu mg\\ x &= \sqrt {\dfrac{{m{v^2} - 2\mu mg}}{k}} ......\left( 1 \right) \end{align*} {/eq}


(1)

Substitute all the values with \mu = 0 in the expression (1) to find the maximum compression of the spring without friction.

{eq}\begin{align*} x &= \sqrt {\dfrac{{\left( {1000\;{\rm{kg}}} \right){{\left( {5\;{\rm{m}}/{\rm{s}}} \right)}^2} - \left( 0 \right)\left( {1000\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m}}/{{\rm{s}}^2}} \right)}}{{6250\;{\rm{N}}/{\rm{m}}}}} \\ x &= 2\;{\rm{m}} \end{align*} {/eq}


Thus, the maximum compression of the spring without friction is {eq}2\;{\rm{m}} {/eq}.


(2)

Substitute all the values with {eq}\mu = 0.5 {/eq} in the expression (1)to find the maximum compression of the spring without friction.

{eq}\begin{align*} x &= \sqrt {\dfrac{{\left( {1000\;{\rm{kg}}} \right){{\left( {5\;{\rm{m}}/{\rm{s}}} \right)}^2} - \left( {0.5} \right)\left( {1000\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m}}/{{\rm{s}}^2}} \right)}}{{6250\;{\rm{N}}/{\rm{m}}}}} \\ x &\approx 1.8\;{\rm{m}} \end{align*} {/eq}


Thus, the maximum compression of the spring with friction is {eq}1.8\;{\rm{m}} {/eq}.


Learn more about this topic:

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Practice Applying Spring Constant Formulas

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Chapter 17 / Lesson 11
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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.


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