A block is placed on a frictionless ramp at a height of 10.5 m above the ground

Question

A block is placed on a frictionless ramp at a height of 10.5 m above the ground Starting from rest, the block slides down the ramp. At the bottom of the ramp, the block slides onto a frictionless horizontal track without slowing down. At the end of the horizontal track, the block slides smoothly onto a second frictionless ramp. How far along the second ramp Hoes the block travel before coming to omentary stop, as measured along ine incline of the ramp? After the block comes to a complete stop on the second ramp, it will then begin moving back down the second ramp. What is the speed of the block when it is 6.75 m, vertically, above the ground?

Explanation / Answer

The block will rise to the same height it started at.

From the diagram, the distance along ramp 2;

D = 10.5/sin20.5° = 29.98 m

In three significant digits, D = 30.0 m

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Apply the law of conservation of energy to find velocity.

The kintic energy is convert in potential energy

1/2 ( m V ^2 ) = mg dy

Here dy = 10.5 - 6.75 = 3.75 m

Solve for V

V = [2*g*dy] = [2*9.8*3.75] = 8.57 m/s

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