a 107 lb little red wagon full of kids is on the top of a 41.4 m hill. the wagon falls down the hill and up a 14.5 m hill. how fast is it going on the top of the second hill if there is no friction?
It falls 41.4 -14.5 meters
so it loses m g h = m g (41.4-14.5) jouless of potential energy
that becomes kinetic energy
so
(1/2) m v^2 = m g (41.4-14.5)
so
v = sqrt [ 2 * 9.81 *(41.4-14.5) ]
@Damon. Question where did the mass go?
To find the speed of the wagon on the top of the second hill, we can make use of the law of conservation of energy. The potential energy at the top of the first hill will be converted into kinetic energy as the wagon moves down the hill and then back into potential energy as it moves up the second hill. Since there is no friction involved, we can assume that the total mechanical energy of the system remains constant.
First, let's calculate the potential energy of the wagon at the top of the first hill:
Potential Energy = mass * gravity * height
mass = 107 lb = 48.54 kg (1 lb = 0.453 kg)
gravity = 9.8 m/s^2 (acceleration due to gravity)
height = 41.4 m
Potential Energy = 48.54 kg * 9.8 m/s^2 * 41.4 m
Next, this potential energy will be converted into kinetic energy as the wagon moves down the first hill. Kinetic energy is given by:
Kinetic Energy = (1/2) * mass * velocity^2
Since the wagon starts from rest at the top of the hill, its initial velocity is 0.
Potential Energy = Kinetic Energy
48.54 kg * 9.8 m/s^2 * 41.4 m = (1/2) * 48.54 kg * velocity^2
Simplifying the equation:
(48.54 kg * 9.8 m/s^2 * 41.4 m) / (1/2 * 48.54 kg) = velocity^2
Now, find the velocity:
velocity = sqrt((48.54 kg * 9.8 m/s^2 * 41.4 m) / (1/2 * 48.54 kg))
The mass of the wagon cancels out, leaving:
velocity = sqrt(9.8 m/s^2 * 41.4 m * 2)
Calculate the velocity:
velocity = sqrt(809.16 m^2/s^2) ≈ 28.45 m/s
Therefore, the speed of the wagon at the top of the second hill, assuming no friction, is approximately 28.45 m/s.