Question

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 1

Answer:

Vo = 4m/s

Explanation:

By conservation of energy:

$1/2*m*Vf^2-1/2*m*Vo^2-m*g*h=0$

Solving for the initial speed:

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

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

Vo=4m/s