Determine its speed after it has slid a distance of 3.00m down the ramp.

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A 2.00kg mass slides down a frictionless ramp inclined at 30 degrees, and the goal is to find its speed after sliding 3.00m. The conservation of energy principles are discussed, with the correct approach involving the work-energy theorem. The user initially calculated the height incorrectly, leading to a speed of 7.14 m/s instead of the expected 5.4 m/s. After clarification, it was noted that the user misapplied trigonometric functions, using CAH instead of SOH to find the height. Correcting this error is essential for accurately determining the speed.
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Homework Statement


A 2.00kg mass starts from rest and slides down a frictionless inclined plane that makes an angle of 30 degrees with the horizontal. Determine its speed after it has slid a distance of 3.00m down the ramp.

Homework Equations


I am leaning about Laws of Conservation of Energy
∑Ek=∑Ep
Ek=(1/2)mv2
Ep=mgh
maybe SOH CAH TOA to get the height

The Attempt at a Solution


using a height of 2.5980, from 3cos(30), i put that into my equation but got 7.1396 m/s
when i know the answer is 5.4 m/s?¿¿
 
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Hey!

You said that you're learning about conservation on Energy,
Then i think it could be usefull the theorem of work and kinect energy.
It simply said that the sum of works that acts on a particle is equal to the change in kinect energy.
When you dou a free body diagram you realize that the only force doing work is the horizontal component of weight,( because the normal force is perpendicular to displacement then work is 0) then you know

mgsin30*3(displacement)=1/2mv^2
you solve for v, and you get 5,4 m/s

Tell me if you already learn the theorem, if your not, there's another way to solve the problem.
 
goracheski said:

Homework Statement


A 2.00kg mass starts from rest and slides down a frictionless inclined plane that makes an angle of 30 degrees with the horizontal. Determine its speed after it has slid a distance of 3.00m down the ramp.

Homework Equations


I am leaning about Laws of Conservation of Energy
∑Ek=∑Ep
Ek=(1/2)mv2
Ep=mgh
maybe SOH CAH TOA to get the height

The Attempt at a Solution


using a height of 2.5980, from 3cos(30), i put that into my equation but got 7.1396 m/s
when i know the answer is 5.4 m/s?¿¿
You've got the right idea, but take a closer look at how you determined the value for h.
 
Thank you gneill I realized I used CAH instead of SOH :S
 
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