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A 0.3 kg block is released from rest after a spring of constant k = 600 N/m has been compressed...

Question:

A 0.3 kg block is released from rest after a spring of constant k = 600 N/m has been compressed 160 mm. The block is not attached to the spring but is simply pushed against it. The block travels in a vertical plane when released. Determine the speed of the block as it passes point B and the force exerted by the loop ABCD on the block as it passes through point B. Assume no friction.

Normal Acceleration:

Normal acceleration is the acceleration which comes in the picture due to change in the direction of velocity. Its direction always acts towards the center of the circle.

Answer and Explanation:


Given Data

  • The mass of the block is :{eq}m = 0.3\;{\rm{kg}} {/eq}
  • The spring constant is: {eq}k = 600\;{\rm{N/m}} {/eq}
  • The length up to which spring is compressed is :{eq}x = 160\;{\rm{mm}} = {\rm{0}}{\rm{.16}}\;{\rm{m}} {/eq}
  • The distance of center of block from the center of block is :{eq}h = 800\;{\rm{mm}} = {\rm{0}}{\rm{.8}}\;{\rm{m}} {/eq}


Apply conservation of energy to find the velocity of block just after it released by the spring.

{eq}\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}k{x^2} {/eq}


Substitute all the values in the above equation.

{eq}\begin{align*} \dfrac{1}{2}\left( {0.3\;{\rm{kg}}} \right)v_1^2 &= \dfrac{1}{2}\left( {600\;{\rm{N/m}}} \right){\left( {0.\;16{\rm{m}}} \right)^2}\\ \left( {0.3\;{\rm{kg}}} \right)v_1^2 &= 192\;{\rm{N}} \cdot {\rm{m}}\left( {\dfrac{{{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}}}{{{\rm{N}} \cdot {\rm{m}}}}} \right)\\ {v_1}& = 7.15\;{\rm{m/s}} \end{align*} {/eq}


Apply conservation to find the velocity of block at point B.

{eq}k.{E_1} = k.{E_2} + P.{E_2} {/eq}

Here, kinetic energy of block at point D is kinetic energy of block at point B is {eq}k.{E_2} {/eq} and potential energy of block at point D is {eq}P.{E_2} {/eq}.


Substitute all the expression in the above equation.

{eq}\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mg\left( {2h} \right) {/eq}

Here, velocity at B is {eq}{v_2} {/eq}.


Substitute all the values in the above equation.

{eq}\begin{align*} \dfrac{1}{2}\left( {0.3\;{\rm{kg}}} \right){\left( {7.15\;{\rm{m/s}}} \right)^2} &= \dfrac{1}{2}\left( {0.3\;{\rm{kg}}} \right)v_2^2 + \left( {0.3\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {2\left( {0.8\;{\rm{m}}} \right)} \right)\\ {v_2}& = 4.44\;{\rm{m/s}} \end{align*} {/eq}


Thus the velocity of block at point B is {eq}4.44\;{\rm{m/s}} {/eq}.


The figure below shows the free body diagram of block at point B.

Free Body Diagram


The normal acceleration of block at B point is,

{eq}{a_n} = \dfrac{{v_2^2}}{h} {/eq}


Substitute all the values in the above equation.

{eq}\begin{align*} {a_n} &= \dfrac{{{{\left( {4.44\;{\rm{m/s}}} \right)}^2}}}{{0.8\;{\rm{m}}}}\\ {a_n} &= 24.64\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}


The weight of the block is,

{eq}W = mg {/eq}


Substitute all the values in the above equation.

{eq}\begin{align*} W &= 0.3\;{\rm{kg}} \times \left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ W&= 2.943\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\left( {\frac{{\rm{N}}}{{{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ W&= 2.943\;{\rm{N}} \end{align*} {/eq}


Apply Newton's second law.

{eq}N + W = m{a_n} {/eq}

Here, {eq}N {/eq} is the normal force applied by the loop on the block.


Substitute all the values in the above equation.

{eq}\begin{align*} N +2.943\;{\rm{N}} & = \left( {0.3\;{\rm{kg}}} \right)24.64\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\ N &= 4.44\;{\rm{N}} \end{align*} {/eq}


Thus normal force applied by the loop on the block is {eq}4.44\;{\rm{N}} {/eq}.


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