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A block is placed on a frictionless ramp at a height of 12.5 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 does the block travel before coming to a momentary stop, as measured along the 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 8.75 m, vertically, above the ground?

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3 answers

  1. initial potential and total energy = E =m g h

    there is no friction in this problem so the total mechanical energy stays the same

    (1/2) m v^2 + m g h = E
    at every height h, you can calculate v

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  2. We do not know the mass of block. The angle of the ramp going down is 44.3 degrees and after it goes across the horizontal surface, it goes up a second ramp at 24.5 degrees

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  3. The second part of the problem can be solved through the following method:
    Ki+Ui=Kf+Uf, which indicates that the inital KE+initial PE=final KE+ final PE.
    Because there is no inital KE or final PE, those values will both be zero. You are then left with just mgh=1/2 mv^2. You can cancel out the m from both sides. Then plug in 8.75m for the h and you will get the velocity.

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