A man skis down from the top of a hill 1500.0 m high, starting from rest. If he loses 50% of his initial mechanical energy to friction during his descent, what is his speed in m/s at the base of the hill?
To find the speed of the man at the base of the hill, we can make use of the principle of conservation of mechanical energy. According to this principle, the total mechanical energy at the top of the hill (potential energy) is equal to the total mechanical energy at the bottom of the hill (kinetic energy).
The potential energy (PE) at the top of the hill is given by the formula:
PE = m * g * h
where m is the mass of the skier, g is the acceleration due to gravity, and h is the height of the hill.
The kinetic energy (KE) at the base of the hill is given by the formula:
KE = (1/2) * m * v^2
where v is the speed of the skier at the base of the hill.
Since the skier loses 50% of his initial mechanical energy to friction, the kinetic energy at the base of the hill will be half of the potential energy at the top of the hill:
(1/2) * PE = KE
Substituting the formulas for potential energy and kinetic energy, we get:
(1/2) * m * g * h = (1/2) * m * v^2
The mass (m) cancels out:
g * h = v^2
To find the speed (v), we can take the square root of both sides:
v = sqrt(g * h)
Substituting the known values, with the acceleration due to gravity being approximately 9.8 m/s^2 and the height of the hill being 1500.0 m:
v = sqrt(9.8 * 1500.0)
Simplifying:
v = sqrt(14700)
Calculating the square root, we find:
v ≈ 121.21 m/s
Therefore, the speed of the man at the base of the hill is approximately 121.21 m/s.