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# A child slides down a frictionless slide of inclination angle $45^\circ$ from a height of $1.8 m$. How fast will she be going at the bottom of the slide? $\left( {g{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 2}}} \right)$

Last updated date: 22nd Jun 2024
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Hint: To find how fast will the child be going at the bottom of the slide, we can compare the height of the slide with kinetic energy.

The child is moving at the bottom of the slide. By using conservation of energy we have,
At the top of the slide, the child has all gravitational potential energy, given by
${U_i} = mgh$
where $m$ is the mass of the child
$g = 9.8m/{s^2}$
The height of the slide is given by
$h = L\sin 45^\circ$
At the bottom of the slide, all of that energy will have been converted to kinetic energy:
$K_f = \dfrac{1}{2}M{V^2}$
we aren't losing any energy to friction, we must have
$\Rightarrow Mg L\sin 45^\circ = \dfrac{1}{2}M{V^2}$
$\Rightarrow L = 3.2$
By solving the above equation we can get the velocity as
$\Rightarrow v{\text{ }} = {\text{ }}6.7{\text{ }}m/s$ .

$KE = \dfrac{1}{2}M{V^2}$