To solve these problems, we can use the principle of conservation of mechanical energy, which states that the sum of the kinetic energy (KE) and potential energy (PE) of an object remains constant as long as only conservative forces (like gravity) are acting on the object.
Let's consider the given information:
Mass of the cyclist and the bicycle (m) = 75 kg
Height of the hill (h) = 15 m
Initial speed of the cyclist (u) = 0 m/s (since she comes to a complete stop)
Acceleration due to gravity (g) = 9.8 m/s^2
a) To calculate the speed at the bottom of the hill, we can equate the potential energy at the top to the kinetic energy at the bottom:
PE (at the top) = KE (at the bottom)
PE = mgh
KE = (1/2)mv^2
where v is the speed at the bottom of the hill.
Plugging in the values, we have:
mgh = (1/2)mv^2
Simplifying, we get:
gh = (1/2)v^2
v^2 = 2gh
v = sqrt(2gh)
Substituting the given values, we have:
v = sqrt(2 * 9.8 * 15)
v ≈ 17.15 m/s
Therefore, the speed of the cyclist at the bottom of the hill is approximately 17.15 m/s.
b) If the cyclist continues pedaling her bike at a constant speed of 5 m/s to the bottom of the hill, her speed will remain the same throughout the descent, since she is providing a constant force to counterbalance any changes caused by gravity.
Therefore, her speed at the bottom of the hill will be 5 m/s.
c) If the hill were twice as steep, the calculation for the speed at the bottom of the hill would be the same as in part a), using the new height of the hill.
Let's denote the height of the steeper hill as 2h.
v = sqrt(2gh)
v = sqrt(2 * 9.8 * (2 * 15))
v ≈ sqrt(2 * 9.8 * 30)
v ≈ 24.28 m/s
Therefore, if the hill were twice as steep, the speed of the cyclist at the bottom would be approximately 24.28 m/s.