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A block is released from rest at the top of a hill of height h. If there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed v. The block is released from rest at the top of another hill with a rough surface and height h. If one-half of the initial mechanical energy of the block-Earth system is lost due to friction as the block descends the hill, the block will reach the bottom of the hill with a speed of

Respuesta :

Answer:

v₁ =√2gh,      v₂ = v₁ /√2

Explanation:

Let's use the concepts of energy and work to analyze each case

hill without rubbing. Energy is conserved

     

starting point. Highest part

         Em₀ = U = mg h

final point. Lower part

         [tex]Em_{f}[/tex] = K = ½ m v²

         Em₀ = Em_{f}

         m g h = ½ m v²

         v₁ =√2gh

rubbing hill

in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction

        W = Em_{f} - Em₀

they indicate half of the initial mechanical energy is lost due to friction

          W = ½ Em₀

we substitute

          - ½ Em₀ = Em_{f} - Em₀

The negative sign is because the friction work always opposes the movement

         Em_{f} = ½ Em₀

        ½ m v₂² = ½ m g h

         v₂ = √½    √2gh

         v₂ = v₁ /√2

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