What is the speed of the crate when it reaches the bottom?

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SUMMARY

The speed of a 20.0-kg crate sliding down a frictionless inclined plane, 3.00 meters high and 20.0 meters long, is calculated using the principle of conservation of energy. The final speed at the bottom of the incline is 7.67 m/s, derived from the equation V = sqrt(2*g*H), where g is the acceleration due to gravity (9.81 m/s²) and H is the height (3.00 m). The mass of the crate does not affect the final speed due to the conservation of energy principle.

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Homework Statement


A 20.0-kg crate slides down an inclined plane that is 3.00-m high and 20.0-m long. If friction is negligible, what is the speed of the crate when it reaches the bottom of the plane?



Homework Equations


W= F*D
K= 1/2 m*(v^2)


The Attempt at a Solution


They don't give me an angle so I'm not sure how I'm supposed to go about solving for the force parallel to the inclined plane. Once I have that I'll be able to finish out the problem. Help please!
(The answer turns out to be 7.67 m/s; I just need to know how to get there)
 
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Just use conservation of energy. You have gravitational potential energy in the beginning. Don't bother with the work W=FD.
 
Ahhh, ok I see. Thanks a lot N-gin!
 
The 20 M long is not needed for this problem because the friction is assumed to be zero. No energy is lost in friction and so M*g*H = (1/2)*M*V^2

So V = sqrt(2*g*H)

So the velocity is independent of the Mass!

If you had friction, then still mass is irrelevant.
 

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