# A 2000 kg car rolls 50 meters down a frictionless 10 degrees incline. If there is a horizontal...

## Question:

A 2000 kg car rolls 50 meters down a frictionless 10{eq}^\circ {/eq} incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the car in a distance of 1 m.

## Spring Constant:

The spring constant tells about the amount of stiffness present in the spring. The spring constant is the practical value that determines the ability of the spring to resist specific deformation under the applied magnitude of the load.

Given data:

• The mass of the car is {eq}m = 2000\,{\rm{kg}} {/eq}
• The given distance is {eq}d = \,50\,{\rm{m}} {/eq}
• The angle of the inclination is {eq}\theta = {10^\circ } {/eq}
• The spring compression is {eq}x = 1\,{\rm{m}} {/eq}

The expression for the work done on the cart due to the gravity is given as

{eq}\begin{align*} W &= mgh\\ &= mgd\sin \theta \end{align*} {/eq}

Substitute the values in the above expression as

{eq}\begin{align*} W &= \left( {2000} \right)\left( {9.8} \right)\left( {50} \right)\sin {10^\circ }\\ & = 1.7 \times {10^5} \end{align*} \cdot\cdot\cdot{(1)} {/eq}

The expression for the work done by the spring is given as

{eq}{W_s} = \dfrac{1}{2}k{x^2} {/eq}

• Here k is the spring constant

Substitute the values in the above expression as

{eq}\begin{align*} {W_s} &= \dfrac{1}{2} \times k \times {\left( 1 \right)^2}\\ &= 0.5k \end{align*} \cdot\cdot\cdot{(2)} {/eq}

Equate equations (1) and (2) as,

{eq}\begin{align*} W &= {W_s}\\ \left( {1.7 \times {{10}^5}} \right) &= 0.5k\\ k &= 3.4 \times {10^5}\,{\rm{kN/m}}\, \end{align*} {/eq}

Thus the spring constant is {eq}k = 3.4 \times {10^5}\,{\rm{kN/m}} {/eq}

Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.