A cyclist and her bicycle have a combined mass of 75 kg. She comes to a complete stop at the crest of 15 m high hill, then coast down the hill without pedalling.
a) Calculate her speed at the bottom of the hill.
b) Calculate her speed if she does not stop and coast, but continues pedalling her bike at 5 m/s to the bottom.
c) What would happen to her speed at the bottom of the hill if it were twice as steep?
m g h = (1/2) m v^2
so
v^2 = 2 g h = 2 * 9.81 * 15 = 294.3
v = 17.2 m/s
Well, if she keeps going 5 m/s then her speed does not change but I think it means she has initial speed of 15 V
kinetic energy at 15 = (1/2) m (225)
so total energy at base = (1/2) m (225) + m g h
so
(1/2) v^2 = 225/2 + g h
v^2 =225 + 294.3 = 519.3
v = 22.8 m/s
nothing if height the same.
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.
To calculate the speed of the cyclist at the bottom of the hill, we can use the principle of conservation of mechanical energy. The total mechanical energy at the top of the hill equals the total mechanical energy at the bottom of the hill, neglecting any energy losses due to friction or air resistance.
a) To calculate the speed when coasting down the hill without pedaling, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill.
Potential Energy at the top = mgh
Kinetic Energy at the bottom = 0.5mv^2
Where:
m = mass of the cyclist + bicycle = 75 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the hill = 15 m
v = speed at the bottom of the hill (what we want to calculate)
Equating the potential energy to the kinetic energy:
mgh = 0.5mv^2
We can cancel out the mass 'm' from both sides of the equation:
gh = 0.5v^2
Rearranging the equation to solve for 'v':
v^2 = 2gh
Taking the square root of both sides of the equation:
v = sqrt(2gh)
Plugging in the values:
v = sqrt(2 * 9.8 * 15)
v ≈ 17.15 m/s
Therefore, the cyclist's speed at the bottom of the hill when coasting is approximately 17.15 m/s.
b) If the cyclist continues pedaling her bike at 5 m/s to the bottom of the hill, we need to consider the additional energy provided by pedaling. In this case, the total mechanical energy at the top of the hill would include both the potential energy and the kinetic energy due to pedaling.
Potential Energy at the top = mgh
Kinetic Energy due to pedaling = 0.5mv^2_p
Where:
v_p = speed due to pedaling = 5 m/s
Equating the total mechanical energy at the top to the kinetic energy at the bottom, including the energy due to pedaling:
mgh + 0.5mv_p^2 = 0.5mv^2
Canceling out 'm' from both sides of the equation:
gh + 0.5v_p^2 = 0.5v^2
Rearranging the equation to solve for 'v':
v^2 = 2(gh + 0.5v_p^2)
Taking the square root of both sides of the equation:
v = sqrt(2(gh + 0.5v_p^2))
Plugging in the values:
v = sqrt(2(9.8 * 15 + 0.5 * 5^2))
v ≈ 20.04 m/s
Therefore, the cyclist's speed at the bottom of the hill when pedaling at 5 m/s is approximately 20.04 m/s.
c) If the hill were twice as steep, we can calculate the effect on the speed at the bottom of the hill. Let's assume the hill's height is now 2 * 15 = 30 m.
Using the same formula as in part (a), we can calculate the speed:
v = sqrt(2 * 9.8 * 30)
v ≈ 24.28 m/s
Therefore, if the hill were twice as steep, the cyclist's speed at the bottom of the hill would be approximately 24.28 m/s.