A 0.0026kg0.0026⁢k⁢g block is in a bowl whose lip is 1.0m1.0⁢m above the ground. At the top, the block has a downward velocity and slides, without friction, down the sloped slide. At the bottom, it has a velocity of 6.0m/s6.0⁢m/s. What is the block’s velocity at the top of the bowl?

Answer :

lcmendozaf

Answer:

Vo = 4m/s

Explanation:

By conservation of energy:

[tex]1/2*m*Vf^2-1/2*m*Vo^2-m*g*h=0[/tex]

Solving for the initial speed:

[tex]Vo = \sqrt{2*(Vf^2/2-g*h)}[/tex]

[tex]Vo = \sqrt{2*(6^2/2-10*1)}[/tex]

Vo=4m/s

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