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.
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.
1 answer
1 answer