Question #cd757

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Question #cd757

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Answer 1 · 2018-02-22T11:51:28

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 $\frac{1}{2} \times 0.25 \times 5^{2} = 3.125 J$ $1/2 ×0.25×5^2=3.125J$ (using $\frac{1}{2} m v^{2}$ $1/2mv^2$ )

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

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

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

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

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

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Question #cd757

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