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A block of mass 4.71 \space kg is attached to a horizontal spring with spring constant k = 3.90...

Question:

A block of mass {eq}4.71 \space kg {/eq} is attached to a horizontal spring with spring constant {eq}k = 3.90 \times 10^2 \space N/m {/eq}. The surface the block rests upon is frictionless. The block is pulled out to {eq}x_i = 0.0520 \space m {/eq} and released.

a. Find the speed of the block at the equilibrium point.

b. Find the speed when {eq}x = 0.027 \space m {/eq}.

c. Repeat part a if friction acts on the block, with coefficient {eq}\mu_k = 0.180 {/eq}.

Friction Force

It is the force that always tries to stop the sliding motion between the interacting body. the friction force is always opposite to the direction of motion. Friction plays a very important role in daily life, because of friction we can walk on the earth.

Answer and Explanation:

Given;

  • Mass of the block {eq}m = 4.71\;{\rm{kg}} {/eq}
  • Spring constant {eq}k = 3.9 \times {10^2}\;{\rm{N/m}} {/eq}
  • Initial displacement {eq}{x_i} = 0.052\;{\rm{m}} {/eq}

a)

Apply energy conservation for finding the velocity,

{eq}\dfrac{1}{2}m{v^2} = \dfrac{1}{2}k{x_i}^2 {/eq}

Substitute the value in above expression,

{eq}\begin{align*} v &= {x_i}\sqrt {\dfrac{k}{m}} \\ v &= 0.052\sqrt {\dfrac{{3.9 \times {{10}^2}}}{{4.71}}} \\ v &= 0.473\;{\rm{m/s}} \end{align*} {/eq}

b)

At the position of x the formula for the velocity is,

{eq}v = \sqrt {\dfrac{k}{m}\left( {{x_i}^2 - {x^2}} \right)} {/eq}

Substitute the value in above expression,

{eq}\begin{align*} v &= \sqrt {\dfrac{{390}}{{4.71}}\left( {{{0.052}^2} - {{0.027}^2}} \right)} \\ v &= 0.4043\;{\rm{m/s}} \end{align*} {/eq}

Speed at {eq}0.027\;{\rm{m}} is 0.4043\;{\rm{m/s}}{/eq} .

c)

Again apply conservation of energy,

{eq}\dfrac{1}{2}m{v^2} + {\mu _k}mg \times {x_i} = \dfrac{1}{2}k{x_i}^2 {/eq}

Substitute the value in above expression,

{eq}\begin{align*} \dfrac{1}{2} \times 4.71 \times {v^2} + 0.18 \times 4.71 \times 9.81 \times 0.052 &= \dfrac{1}{2} \times 390 \times {0.052^2}\\ v &= 2.03\;{\rm{m/s}} \end{align*} {/eq}

So the speed of the block when friction act is {eq}2.03\;{\rm{m/s}}{/eq} .


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
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