A roller-coaster car rolls down a track, reaching a speed of v0 at the bottom. If the height of the track is increased by a factor of 2.00, assuming the track is frictionless, by what factor will the speed at the bottom increase?
To determine how the speed at the bottom of the track changes when the height is increased by a factor of 2.00, we can apply the principle of conservation of energy.
The initial energy of the roller coaster at the top of the track is given by its potential energy, which is proportional to its height, and its kinetic energy, which is proportional to the square of its velocity.
According to the conservation of energy, the total initial energy will be equal to the total final energy. In this case, the initial energy is given by the potential energy at the top, which is mgh, where m is the mass of the roller coaster, g is the acceleration due to gravity, and h is the initial height of the track. The initial velocity can be assumed to be zero.
The final energy is given by the potential energy at the increased height, which is 2mgh, and the kinetic energy at the bottom, which is (1/2)mv^2, where v is the final velocity at the bottom of the track.
Setting the total initial energy equal to the total final energy, we have:
mgh = 2mgh + (1/2)mv^2
Canceling the m and h terms and rearranging the equation, we get:
0 = (1/2)mv^2 - mgh
Simplifying further, we have:
(1/2)v^2 = gh
Dividing both sides by g and taking the square root, we get:
v = sqrt(2gh)
This equation shows that the velocity v is proportional to the square root of the product of the gravitational acceleration g and the initial height h.
Now, if we increase the height of the track by a factor of 2.00, the new height becomes 2h. Plugging this value into the equation, we get:
v_new = sqrt(2g(2h))
Simplifying further, we have:
v_new = sqrt(4gh) = 2*sqrt(gh)
Comparing this to the original velocity v, we see that the new velocity is twice the original velocity:
v_new = 2v
Therefore, when the height of the track is doubled, the speed at the bottom will increase by a factor of 2.00.