|Mar18-11, 08:42 PM||#1|
work done with cyclist
1. The problem statement, all variables and given/known data
a cyclist rides up a steady slope for a distance of 1200m and rises 90m vertically as a result. the total mass of rider + cycle is 80kg
a) assume a drag force of 10N opposing the motion(due to air resistance and friction), what Total Work must be output by the rider for this climb?
b) if the efficiency for riding in this situation is 20%, what energy must be metabolised for this ride? How much pure oxygen at standard atmospheric pressure is used in metabolising this energy?
2. Relevant equations
Work done = F d cos (Theta)
total work done = W(normal) + W(fr) + W(applied) + W(gravitational)
3. The attempt at a solution
hence cyclist moves at constant speed, W(normal) is 0
distance of the slope is root(1200^2 + 90^2) = about 1203.4
W(fr) = 10N * 1203.4 = 12034 J
and W(g) = 80 * 9.8 * 90 = 70560 J
|Mar18-11, 09:22 PM||#2|
First of all, just to clarify...
The total work done will be the work done in overcoming the drag force (10N), weight (mg), and normal force (F_N). So the total force (depending on which direction you say is positive or negative) should be mg + F_N + 10N. Say you align the axes with the slope. Since the y-component of the forces cancels out, it's best to look at the x-components. You should be able to figure this out using sin theta, from the lengths given. Also, I would think that the slope is the hypoteneuse of a right triangle, therefore the other side would instead be square root of (1200^2-90^2).
Efficiency is simply the work you get out over the work you put in, therefore since you know the work you must get out and the efficiency it should be simple enough to get the work you must put in with a little algebra.
However, I'm not sure about the oxygen problem- do you have a value for rate of oxygen metabolism? Once you do, it will be fairly easy to get the final answer.
Hope this helps :)
|Mar19-11, 04:24 AM||#3|
what do you mean by cancel y component???
isn't normal force is perpendicular to the surface and mg is straight down??
how can these be cancelled?
|Mar19-11, 10:19 AM||#4|
work done with cyclist
Pehaps I phrased my answer a little vaguely. What I meant to say is that the normal force is equal to the y-component of the weight. Thus the y-components cancel out. Take a look at the attached free-body diagram and see if that makes sense.
|Mar26-11, 11:26 PM||#5|
Im still struggling with this, could someone explain it again?
|Mar26-11, 11:28 PM||#6|
I know exactly where you get this question from :)
It says rides up a slope, so that means that the distance of the hypotenuse is the slope, which is 1200m.
So just find the loss of kinetic energy in the ride, which is equal to the work. He has to overcome of the loss of kinetic energy to gravitational energy, which is 80*90*9.8=70560J
Then he has to overcome the drag, and work due to that is w=Fd, Wdrag=10*1200=12000J
therefore, total work is 70560+12000=82560J.
Not sure about the rest though
|Mar27-11, 02:06 AM||#7|
in relation to the question....
if this ride was done at an altitude where atmospheric pressure was 0.90 atmosphere, how much air would be inhaled?
|cyclist, slope, work done|
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