A cyclist approaches the bottom of a gradual hill at a speed of 20 m/s. The hill is 4.4 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.
To find the speed at which the cyclist crests the hill, we will use the principle of conservation of mechanical energy. The total mechanical energy of the cyclist on the hill is equal to the sum of the potential energy and kinetic energy.
The potential energy of the cyclist at the bottom of the hill is given by the formula:
PE = mgh
Where m is the mass of the cyclist, g is the acceleration due to gravity, and h is the height of the hill.
The kinetic energy of the cyclist at the bottom of the hill is given by the formula:
KE = (1/2)mv^2
Where v is the initial speed of the cyclist at the bottom of the hill.
Since the cyclist is coasting, there is no external work being done on the system, and thus the total mechanical energy remains constant throughout the motion. Therefore, the total mechanical energy at the bottom of the hill is equal to the total mechanical energy at the crest of the hill.
At the top of the hill, the potential energy is given by:
PE' = mgh'
Where h' is the height of the hill at the crest.
And the kinetic energy at the top of the hill is given by:
KE' = (1/2)mv'^2
Where v' is the speed of the cyclist at the crest of the hill.
Since the total mechanical energy is conserved, we can write:
PE + KE = PE' + KE'
mgh + (1/2)mv^2 = mgh' + (1/2)mv'^2
Since the mass of the cyclist cancels out, we can simplify the equation to:
gh + (1/2)v^2 = gh' + (1/2)v'^2
Solving for v' (the speed at which the cyclist crests the hill), we get:
v' = sqrt(v^2 + 2g(h - h'))
Plugging in the given values, we have:
v' = sqrt(20^2 + 2 * 9.8 * (4.4 - 0))
Simplifying further,
v' = sqrt(400 + 2 * 9.8 * 4.4)
v' = sqrt(400 + 86.24)
v' = sqrt(486.24)
v' ≈ 22.04 m/s
Therefore, the speed at which the cyclist crests the hill is approximately 22.04 m/s.
The initial kinetic energy (1/2)M V^2 must equal or exceed the potential energy change at the top of the "hill", which is M g H.
Therefore V > sqrt (2 g H)