# An object moving along a horizontal track collides with and compresses a light spring (which...

## Question:

An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is {eq}38.6\, N/m {/eq}, the mass of the object {eq}0.270\, kg {/eq}, and the speed of the object is {eq}1.20\, m/s {/eq} immediately before the collision.

(a) Determine the spring's maximum compression if the track is frictionless.

(b) If the track is not frictionless, will the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)?

## Spring Force:

Spring force is the amount of force available in the spring due to the compression or extension of the spring. It is related to the spring constant, and the amount of extension and compression of spring and direction of force is always opposite to the movement.

## Answer and Explanation: 1

** Given data **

- The spring constant of the spring is: {eq}k = 38.6\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} {/eq}

- The mass of the object is: {eq}m = 0.270\;{\rm{kg}} {/eq}

- The speed of the object is: {eq}v = 1.20\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} {/eq}

** (a) **

The work done by the spring is converted in to kinetic energy of the object. The expression for the kinetic energy and work done is given as:

{eq}\begin{align*} KE &= W{D_s}\\ \dfrac{1}{2}m{v^2} &= \dfrac{1}{2}k{x^2}\\ x &= v \times \sqrt {\dfrac{m}{k}} \end{align*} {/eq}

Substitute the value in the above equation.

{eq}\begin{align*} x &= 1.20 \times \sqrt {\dfrac{{0.270}}{{38.6}}} \\ x &= 0.1\;{\rm{m}} \end{align*} {/eq}

Thus, the spring's maximum compression if the track is frictionless is {eq}0.1\;{\rm{m}} {/eq}.

** (b) **

The summation of the work done by the spring and frictional force on the object is equal to the kinetic energy of an object. The expression is given as:

{eq}\begin{align*} W{D_s} + W{D_f} &= KE\\ \dfrac{1}{2}k{x^2} + \mu \times mgx &= \dfrac{1}{2}m{v^2} \end{align*} {/eq}

Where, {eq}W{D_s} {/eq} is the work done by the spring, {eq}W{D_f} {/eq} is the work done due to the frictional force, {eq}KE {/eq} is the kinetic energy of an object, and x is the maximum compression of the spring and the maximum coefficient of friction is: {eq}\mu = 0.5 {/eq}.

Substitute the value in the above equation.

{eq}\begin{align*} \dfrac{1}{2} \times 38.6 \times {x^2} + 0.5 \times 0.27 \times 9.81 \times x &= \dfrac{1}{2} \times 0.270 \times {1.2^2}\\ 19.3{x^2} + 1.32x - 0.194 &= 0\\ x &= \dfrac{{ - 1.32 \pm \sqrt {{{1.32}^2} - 4 \times 19.3 \times \left( { - 0.194} \right)} }}{{2 \times 19.3}}\\ x &= 0.071\;{\rm{m}} \end{align*} {/eq}

Thus, the value of the maximum compression of the spring when the maximum friction is present in the surface is {eq}0.071\;{\rm{m}} {/eq}.

Thus, If the track is not frictionless, the spring's maximum compression be less than the value obtained in part (a).

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Chapter 4 / Lesson 19After 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.