A ball rolls without slipping down the track as shown in the figure below, starting from rest at a height h1 = 27 m as shown. The ball is traveling horizontally when it leaves the bottom of the track, which has a height h2 = 8.0 m.
To find the speed of the ball when it leaves the bottom of the track, you can use the principle of conservation of mechanical energy.
The principle of conservation of mechanical energy states that the total mechanical energy of a system remains constant if no external forces are acting on it. In this case, the ball only experiences the force of gravity, which is an internal force, so mechanical energy is conserved.
The total mechanical energy of the ball at the initial position (height h1) is given by the sum of its potential energy and kinetic energy:
E1 = mgh1 + 0, where m is the mass of the ball, g is the acceleration due to gravity, and h1 is the initial height.
The total mechanical energy of the ball at the final position (height h2) is given by:
E2 = mgh2 + 0, where h2 is the final height.
Since energy is conserved, E1 = E2.
Setting the two equations equal to each other, we get:
mgh1 = mgh2
Canceling out the mass term, we have:
gh1 = gh2
Now, we can solve for the speed of the ball when it leaves the bottom of the track. The potential energy at height h1 is converted into kinetic energy at height h2.
To find the speed, we can use the equation for potential energy:
Potential energy = mgh
Setting the potential energy at height h1 equal to the kinetic energy at height h2, we have:
mgh1 = 1/2 mv^2
where v is the velocity or speed of the ball when it leaves the bottom of the track.
To find the speed, we can rearrange the equation:
v^2 = 2gh1
Taking the square root of both sides, we get:
v = sqrt(2gh1)
Substituting the values h1 = 27 m and g = 9.8 m/s^2, we can calculate the speed of the ball when it leaves the bottom of the track:
v = sqrt(2 * 9.8 * 27)
v ≈ 26.87 m/s
So, the speed of the ball when it leaves the bottom of the track is approximately 26.87 m/s.