Question

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? \[\left( {g{\text{ }} = {\text{ }}9.8{\text{ }}m{s^{ - 2}}} \right)\]

Answer 1

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.

Complete step by step answer:
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
\[\Rightarrow v{\text{ }} = {\text{ }}6.7{\text{ }}m/s\] .

Additional information:
Kinetic energy is directly proportional to the mass of the object and the square of its velocity:
$KE = \dfrac{1}{2}M{V^2}$
If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.
It is the energy that an object possesses because of its motion. Kinetic energy is dependent on the velocity of the object squared. And this energy should be either a zero or a negative value.

Note: This kinetic energy refers to the work required to accelerate the body of a particular mass from a state of rest to its stated velocity. The standard unit of this kinetic energy happens to be the joule.