A child of mass m rides on an irregularly curved slide of height h = 2m. The child starts from...

Question:

A child of mass {eq}m {/eq} rides on an irregularly curved slide of height {eq}h = 2m {/eq}. The child starts from rest at the top of the slide.

a. What is the speed of the child at the bottom of the slide assuming that no friction is present (the slide is wet)?

b. If a force of kinetic friction acts on the child, how much mechanical energy does the system lose?

Assume that {eq}v_f = 3.0 m/s {/eq} and that {eq}m_{child}= 20kg. {/eq}

Kinetic and Potential Energy:

The kinetic energy of an object comes into play when the object is moving at a certain velocity (magnitude and direction). The potential energy of the system is calculated using the given height of the system from a reference point.


Answer and Explanation:


Given Data

The initial height at which the child is situated is: {eq}H = 2\;{\rm{m}} {/eq}.


a. The potential energy at the top of the slide is:

{eq}PE = {m_{child}}gh {/eq}


The kinetic energy at the bottom of the slide is:

{eq}KE = \dfrac{1}{2}{m_{child}}{v^2} {/eq}


Here, the speed of the child is {eq}v {/eq}.


As long as there's no friction, according to the conservation of energy:

{eq}\begin{align*} PE &= KE\\ \ \\ {m_{child}}gH &= \dfrac{1}{2}{m_{child}}{v^2}\\ \ \\ v &= \sqrt {2gH} \end{align*} {/eq}


Substituting the known values in the equation:

{eq}\begin{align*} v &= \sqrt {2\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2\;{\rm{m}}} \right)} \\ \ \\ &= 6.264\;{\rm{m/s}} \end{align*} {/eq}


Thus, the speed of the child at the bottom of the slide is {eq}6.264\;{\rm{m/s}} {/eq}.


b. We know that:

  • The final speed of the child is {eq}{v_f} = 3\;{\rm{m/s}} {/eq}.
  • The mass of the child is {eq}{m_{child}} = 20\;{\rm{kg}} {/eq}.


The mechanical energy loss (considering the friction force) is:

{eq}E = PE - \dfrac{1}{2}{m_{child}}v_f^2 {/eq}


Substitute the known values in the equation:

{eq}\begin{align*} E &= {m_{child}}gH - \dfrac{1}{2}{m_{child}}v_f^2\\ \ \\ &= \left( {20\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)2\;{\rm{m}} - \left( {0.5\left( {20\;{\rm{kg}}} \right)} \right){\left( {3\;{\rm{m/s}}} \right)^2}\\ \ \\ &= 302.4\;{\rm{J}} \end{align*} {/eq}


Thus, the mechanical energy loss is {eq}302.4\;{\rm{J}} {/eq}.


Learn more about this topic:

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