How Far Does a Block Slide Up an Inclined Plane?

AI Thread Summary
A 2.3-kg block slides at 5.9 m/s on a frictionless surface before transitioning to a frictionless ramp inclined at 24 degrees. The energy conservation equation used is mgh + 0.5 mv^2 = TE. The initial calculation for height (h) was incorrect due to misapplication of mass in the formula. The correct approach requires recognizing that mass cancels out, leading to the formula h = v^2 / (2g sin 24). The final corrected calculation shows that the block slides up the ramp approximately 3.79 meters before coming to rest.
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[SOLVED] Energy Conservation Problem Please HELP!:)

Homework Statement



A 2.3-kg block slides along a frictionless horizontal surface with a speed of 5.9 m/s After sliding a distance of 6 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 24o to the hori-zontal.
How far up the ramp does the block slide before coming monetarily to rest?

Homework Equations



mgh + .5 mv^2 = TE

The Attempt at a Solution



I tried doing it with this equation, but my answer was incorrect

mgh sin 24 = 1/2mv2
solve for h= v2/(2.3g sin 24 ) = [5.9^2 /(2.3 ´ 9.81 ´ sin 24o)] = 3.793096 m
 
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I figured out my mistake. I forgot that I canceled out my masses and was using the equation h=v^2 / 2gsin...
 
You shouldn't have a "2.3" in the denominator, since the masses cancel. This should be a 2 (from the 1/2 on the RHS).
 
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