A cyclist approaches the bottom of a gradual hill at a speed of 12.4 m/s. The hill is 6.7 m high and the cyclist estimates 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 coasts over the hill?
To find the speed at which the cyclist coasts over the hill, we can use the principle of conservation of mechanical energy. The total mechanical energy of the cyclist remains constant throughout the motion as there are no external forces acting on her.
The total mechanical energy of the cyclist can be expressed as the sum of kinetic energy (KE) and potential energy (PE):
TE = KE + PE
Initially, at the bottom of the hill, the cyclist has only kinetic energy, given by:
KE1 = (1/2) * m * v1^2
Where:
m is the mass of the cyclist
v1 is the initial speed of the cyclist (12.4 m/s)
At the top of the hill, the cyclist has both kinetic and potential energy, given by:
KE2 = (1/2) * m * v2^2
PE = m * g * h
Where:
v2 is the final speed of the cyclist at the top of the hill
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the hill (6.7 m)
Since the mechanical energy is conserved, the sum of the initial kinetic energy and the initial potential energy equals the sum of the final kinetic energy and the final potential energy:
KE1 = KE2 + PE
Substituting the expressions for kinetic energy and potential energy, we have:
(1/2) * m * v1^2 = (1/2) * m * v2^2 + m * g * h
We can cancel out the mass (m) on both sides of the equation:
(1/2) * v1^2 = (1/2) * v2^2 + g * h
Now, let's substitute the given values into the equation:
v1 = 12.4 m/s, h = 6.7 m, g = 9.8 m/s^2
(1/2) * (12.4)^2 = (1/2) * v2^2 + 9.8 * 6.7
153.76 = 0.5 * v2^2 + 65.66
Rearranging the equation to solve for v2:
0.5 * v2^2 = 153.76 - 65.66
v2^2 = (153.76 - 65.66) / 0.5
v2^2 = 176.2
v2 = sqrt(176.2)
Calculating the square root gives:
v2 ≈ 13.27 m/s
Therefore, the speed at which the cyclist coasts over the hill is approximately 13.27 m/s.