Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 4.00 m/s. At what final speed would the child enter the water if the water slide were twice as high? Ignore friction and resistance from the air and the water lubricating the slide.
(1/2) m v^2 = m g h
v = sqrt (2 g h)
or
v = k sqrt(h)
if you double h, you multiply v by sqrt 2
What is k?
To solve this problem, we can use the principle of conservation of energy. The initial potential energy of the child at the top of the slide is converted into kinetic energy at the bottom of the slide.
We know that the child enters the water at a final speed of 4.00 m/s. This means that the total change in kinetic energy equals the total change in potential energy.
Let's calculate the initial and final potential energy of the child for the given water slide height and use that to find the final speed for a two times higher water slide.
1. For the given water slide height:
The final kinetic energy = initial potential energy
Ek(final) = Ep(initial)
(1/2) * m * v^2 = m * g * h
where,
m = mass of the child (we'll assume it cancels out when dividing the two equations)
v = final speed = 4.00 m/s
g = acceleration due to gravity = 9.81 m/s^2 (approximate value on Earth)
h = initial height of the water slide
Rearrange the equation:
v^2 = 2 * g * h
Plug in the given values and calculate the initial height:
4.00^2 = 2 * 9.81 * h
16.00 = 19.62 * h
h = 16.00 / 19.62
h ≈ 0.8153 meters
So, for the given water slide height, the initial height is approximately 0.8153 meters.
2. For two times higher water slide:
Let's denote the new height of the slide as H (twice the initial height).
To find the final speed when entering the water, we'll use the same formula as before:
v'^2 = 2 * g * H
Substituting the known values:
v'^2 = 2 * 9.81 * (2h)
v'^2 = 2 * 9.81 * 2 * 0.8153
v'^2 = 31.43
Taking the square root of both sides:
v' = √(31.43)
v' ≈ 5.60 m/s
Therefore, if the water slide were twice as high, the child would enter the water with a final speed of approximately 5.60 m/s.
To solve this problem, we can use the principle of conservation of energy. At the top of the water slide, the child's initial energy is purely gravitational potential energy, which is given by the formula:
PE_initial = m * g * h
where m is the mass of the child, g is the acceleration due to gravity, and h is the height of the water slide.
As the child slides down the water slide, gravitational potential energy is converted into kinetic energy. At the bottom of the slide, the child's final energy is purely kinetic energy, given by the formula:
KE_final = (1/2) * m * v^2
where v is the final speed of the child entering the water.
Since there is no loss of energy due to friction or resistance, we can equate the initial potential energy to the final kinetic energy:
PE_initial = KE_final
m * g * h = (1/2) * m * v^2
Canceling out the mass from both sides of the equation, we get:
g * h = (1/2) * v^2
Now, let's solve the equation for the final speed, v.
v^2 = 2 * g * h
Taking the square root of both sides, we have:
v = √(2 * g * h)
Given that the child enters the water at a final speed of 4.00 m/s, we can substitute this value into the equation to find the height of the water slide.
4.00 = √(2 * g * h)
Squaring both sides of the equation, we get:
16 = 2 * g * h
Dividing both sides of the equation by 2 * g, we can solve for h:
h = 16 / (2 * g)
Now, if the water slide were twice as high, we can substitute this new height, 2h, into the equation:
v' = √(2 * g * 2h)
Simplifying further:
v' = √(4 * g * h)
v' = 2 * √(2 * g * h)
Therefore, if the water slide were twice as high, the final speed at which the child would enter the water would be two times the original final speed. In this case, it would be:
v' = 2 * 4.00 m/s = 8.00 m/s