How many times as high to achieve twice speed

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To achieve a speed of 2v at the bottom of a frictionless ramp, the height must be four times greater than the height needed to reach speed v. This conclusion is derived from the conservation of energy principle, where potential energy (mgh) is converted to kinetic energy (1/2 mv^2). The discussion clarifies that while it may seem intuitive that doubling the speed would require doubling the height, the relationship between speed and height is quadratic. The final answer confirms that the required height is indeed four times greater, resolving initial confusion. Understanding this concept is crucial for applying the laws of physics effectively.
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A block initially at rest is allowed to slide down a frictionless ramp and attains a speed of v at the bottom. In order to achieve a speed of 2v instead at the bottom, how many times as high must the new ramp be?

What laws does this pertain to? I don't really know where to start.
 
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It pertains to the conservation of energy.
mgh = 1/2*m*v^2.
 
rl.bhat said:
It pertains to the conservation of energy.
mgh = 1/2*m*v^2.

But I don't understand...velocity is only on one side of that equation. So how would you work it out for this problem?
 
Exactly! What you have is the height of an object as a function of it's speed at the end. So if you know you must be a height, h, to achieve a velocity v, what must h be in order to achieve 2v?
 
Pengwuino said:
Exactly! What you have is the height of an object as a function of it's speed at the end. So if you know you must be a height, h, to achieve a velocity v, what must h be in order to achieve 2v?

2h, but it says the answer is 4 times as high...that's what's confusing me...I don't know how to reach this final answer of 4.
 
Nevermind, haha, I got it. Thanks a lot! Sorry for being a bit slow on the uptake : P
 
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