Respuesta :
Answer:
new height of ramp must be 4 times that of the first ramp.
Explanation:
From conservation of energy, we know that;
Potential energy at the top of ramp = kinetic energy at the bottom of ramp.
Thus;
mgh_t = ½mv²
m will cancel out to give;
gh_t = ½v²
Thus means that the height of the first ramp is directly proportional to the square of the speed.
Thus;
h_t ∝ v²
Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.
Thus;
h'_t ∝ (2v)²
h'_t ∝ 4v²
From earlier we saw that;
h_t ∝ v²
Thus;
New height of ramp is;
h'_t ∝ 4h_t
Thus, new height of ramp must be 4 times that of the first ramp.
To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
The given parameters;
- velocity at the bottom ramp, = v
- let the height of the ramp = h
Apply the principle of conservation of energy;
P.E = K.E
[tex]mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}[/tex]
To increase the speed 2v at the bottom of the ramp, the height will be;
[tex]H = \frac{(2v)^2}{2g} \\\\H = \frac{4v^2}{2g} \\\\H = 4(\frac{v^2}{2g} )\\\\H = 4(h)[/tex]
Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
Learn more here:https://brainly.com/question/20626677