# Question cd757

Feb 22, 2018

When the object reaches the bottom of the inclined plane,its total energy is purely kinetic,considering horizontal to be the reference level.

And,the energy value is 1/2 ×0.25×5^2=3.125J (using $\frac{1}{2} m {v}^{2}$)

Now,total energy on top of the inclined plane was purely potential energy as the object was at rest,and its value was $m g h$ (where, $h$ is the height of the plane)

As, $\frac{h}{l} = \sin 30$ so, $h = \frac{l}{2} = \frac{6}{2} = 3 m$ ($l$ is the length of the inclined plane)

So,total energy at the beginning of the journey was 0.25×9.81×3=7.36 J#

So,this decrease in total energy value suggests that it must have been utilized to overcome the frictional force,

And the value is $\left(7.36 - 3.125\right) = 4.24 J$