Two velocities, unknown: height, mass, theta HELP

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SUMMARY

The discussion centers on a physics problem involving a block sliding down a ramp, where the speeds at points A and B are given for two trials. The first trial has speeds of 1.85 m/s at point A and 2.60 m/s at point B, while the second trial has a speed of 3.95 m/s at point A. Using the conservation of mechanical energy equation, gh(initial) + (v(initial)^2)/2 = gh(final) + (v(final)^2)/2, participants conclude that the difference in gravitational potential energy between points A and B remains constant across both trials, leading to the same change in kinetic energy. This insight simplifies the problem, allowing for the determination of the unknown speed at point B in the second trial.

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Homework Statement


A block is sent sliding down a ramp. Its speeds at points A and B are 1.85 m/s and 2.60 m/s, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is 3.95 m/s. What then is its speed at point B?

(point a is logically higher on the ramp than point b)

Homework Equations



The goal of this one is to use conservation of mechanical energy to deduce the final velocity of the second trial. Heres what i have:

since this is clearly a problem that does not need to involve the mass in the mechanical energy equation, i set something up that should describe the motion and the change in kinetic energy of this object:

gh(initial) + (v(initial)^2)/2 = gh(final) + (v(final)^2)/2

This should be set up twice (once for each set of velocities, the second set containing an unknown velocity for vfinal.)

What I'm confused about is how the kinetic energy and work equations come into play if at all. There is clearly no frictional force acting on the block so the acceleration is constant. Where do i go from here? Anybody have a suggestion on how to tackle the heights or weather or not it is necessary to find those quantities? i assume it is but i could very well be wrong.
 
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Consider the difference in gravitational potential (A to B) between the two runs.

Won't that necessarily be the same for both runs?

So doesn't that imply the difference in kinetic energy must also be the same?
 
wow...thats deathly simple. As usual, the hardest problems take the least amount of time... if you know what your doing!
 

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