A cyclist moves along a horizontal road. She pushes on the pedals with a constant power of 250W. The mass of the cyclist and bicycle is 85kg. The total drag force is 0.4V^2, where v is the speed of the cyclist.
i) Calculate the energy provided by the cyclist each minute when the overall efficiency of the cyclists muscles is 65% (in J)
ii) Calculate the drag force and hence the instantaneous acceleration of the cyclist when the speed is 6.0ms-1.
Please help me I had the mark scheme to this but couldn’t understand it so please be clear and write even every single small thing you do and write out full equations with no initials or letters to represent just full words.
i) First, we'll determine the energy provided by the cyclist each minute.
Power (P) is given as energy transferred (E) per unit time (t).
P = E/t
We know that the power is 250 W (Joules per second), and the efficiency is 65%. Therefore, the actual energy provided each second by the cyclist is:
Actual energy (E') = Efficiency * Power
E' = 0.65 * 250 W
E' = 162.5 J/s
Now we want to calculate the energy provided each minute. We can write this as:
Energy provided per minute (E_m) = E'; the energy provided per second * 60 (seconds in a minute)
E_m = 162.5 J/s * 60s
E_m = 9750 J
So, the energy provided by the cyclist each minute is 9750 Joules.
ii) Now let's calculate the drag force when the speed is 6.0 m/s.
The total drag force is given as F_drag = 0.4 * V^2
F_drag = 0.4 * (6.0 m/s)^2
F_drag = 0.4 * 36 N
F_drag = 14.4 N
Now let's find the instantaneous acceleration (a) using Newton's second law:
F_net = m*a
where
F_net = net force acting on the cyclist
m = mass of cyclist and bicycle = 85 kg
a = acceleration
Since the drag force is the only force opposing the cyclist's forward motion, our net force is:
F_net = F_drag
Now we can calculate the acceleration:
a = F_net / m
a = 14.4 N / 85 kg
a = 0.1694 m/s²
Hence, the instantaneous acceleration of the cyclist when the speed is 6.0 m/s is 0.1694 m/s².
To calculate the energy provided by the cyclist each minute, we can first calculate the power output of the cyclist using the given constant power of 250W. Then, we can use the equation for power to calculate the energy provided in one second, and then convert it to minutes.
i)
Step 1: Convert the constant power to power output (P) in Watts.
Given: Constant power (P) = 250W
Step 2: Calculate the power output (P') in Watts (W).
Power output (P') = Efficiency (η) * Constant power (P)
η = 65% = 0.65
P' = 0.65 * 250W
Step 3: Calculate the energy provided in one second (E') in Joules (J).
Energy provided (E') = Power output (P') * Time (t)
Time (t) = 1 second
E' = P' * t
Step 4: Convert the energy provided to minutes.
Since there are 60 seconds in a minute, we can multiply the energy provided in one second by 60.
Energy provided (E) = E' * 60
ii)
To calculate the drag force and the instantaneous acceleration of the cyclist, we can use the given equation for drag force and apply Newton's second law of motion.
Step 1: Calculate the drag force (F) in Newtons (N) using the given equation.
Total drag force (F) = 0.4v^2
Speed (v) = 6.0 m/s
F = 0.4 * (6.0 m/s)^2
Step 2: Calculate the mass (m) of the cyclist and bicycle in kilograms (kg).
Given: Mass (m) = 85 kg
Step 3: Apply Newton's second law of motion to calculate the instantaneous acceleration (a) of the cyclist.
Force (F) = mass (m) * acceleration (a)
a = F / m
Now, let's perform the calculations:
i)
Step 1: Constant power (P) = 250W
Step 2: Power output (P') = 0.65 * 250W = 162.5W
Step 3: Energy provided in one second (E') = P' * t = 162.5W * 1s = 162.5J
Step 4: Energy provided in one minute (E) = E' * 60 = 162.5J * 60 = 9750J
Therefore, the energy provided by the cyclist each minute when the overall efficiency of the cyclist's muscles is 65% is 9750 Joules (J).
ii)
Step 1: Speed (v) = 6.0 m/s
Drag force (F) = 0.4 * (6.0 m/s)^2 = 0.4 * 36.0 m^2/s^2 = 14.4 N
Step 2: Mass (m) = 85 kg
Step 3: Instantaneous acceleration (a) = F / m = 14.4 N / 85 kg = 0.169 m/s^2
Therefore, the drag force is 14.4 Newtons (N), and the instantaneous acceleration of the cyclist when the speed is 6.0 m/s is 0.169 m/s^2.
To answer these questions, let's break them down step by step.
i) To calculate the energy provided by the cyclist each minute, we need to find the power expended by the cyclist and then convert it to energy.
The power expended by the cyclist is given as 250W (watts). Power (P) is defined as the rate at which work is done or energy is consumed. Mathematically, it can be represented as P = W/t, where W is the work done or energy consumed, and t is the time taken.
Since we need to find the energy provided each minute, we have t = 1 minute = 60 seconds.
So, the energy provided is given by E = P × t.
Substituting the values, we get E = 250W × 60s.
Since the overall efficiency of the cyclist's muscles is 65%, we need to take this into account. Efficiency (η) is defined as the ratio of useful output energy to input energy.
η = useful output energy / input energy.
In this case, the input energy is the energy provided by the cyclist, and the useful output energy is the work done against the drag force (as it propels the cyclist forward).
η = W(output) / W(input).
Given the overall efficiency (η = 65%), we can rewrite the equation as:
W(output) = η × W(input).
Now, let's calculate the energy provided by the cyclist each minute, considering the overall efficiency:
E(output) = η × E(input).
Substituting the values, we get E(output) = 0.65 × (250W × 60s).
ii) To calculate the drag force and instantaneous acceleration of the cyclist when the speed is 6.0 m/s, we'll use the formula for drag force (F_drag) and the second law of motion.
The formula for drag force is given as F_drag = 0.4 × V^2, where V is the speed of the cyclist.
So, when the speed is 6.0 m/s:
F_drag = 0.4 × (6.0 m/s)^2.
To find the instantaneous acceleration (a), we use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F_net = m × a).
In this case, the net force is the difference between the force applied by the cyclist (which is the power divided by velocity) and the drag force:
F_net = P / V - F_drag.
Since we need to find the instantaneous acceleration, we can rearrange the equation:
a = (P / V - F_drag) / m.
Substituting the values, we get:
a = (250W / 6.0 m/s - (0.4 × (6.0 m/s)^2)) / 85kg.
Now, you can plug in these values into a calculator to find the results.