When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.45 m² and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90 . Use 1.2 kg/m³ as the density of air at room temperature. (c) The food calorie is equivalent to 4190 J. How many calories does the cyclist burn if he rides over level ground at 7.3 m/s for 1 h?

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Accepted Answer

Hey, everyone. So this problem starts with reminding us that air exerts a significant drag force on moving objects. A motorcycle rider needs to keep the engine on when riding at a constant speed on a flat surface to counter the drag effects. Presume that all of the engines, mechanical energy output is converted to work against the drag. We're also told to assume that the rider and motorcycle has a cross sectional area of 0.8 m per meter squared and a drag coefficient Of 0.85 is due to the dynamic inefficiency. We're told to take the density of air to be a typical 1.2 kg per meter cubed and the engine and transmission to be 31% efficient at converting chemical energy to mechanical energy. We are then asked to determine the chemical energy intake of the engine and units of kilocalories. And they give us that conversion between kilocalories and killer jewels of one killer calorie equals 4.19 killer jewels. We were asked to determine that intake when the rider travels or at 50 km km for 40 minutes. Okay. So the first thing we're gonna do here is draw our free body diagram. So we know we have the force of the engine moving the motorcycle forward and we have the drag force acting in the direction opposite of motion. So then the negative X direction from Newton's second law, we can take some of forces in the X direction to be equal to mass times acceleration. And because we know that we are working that we are Traveling at a constant speed of 50 km/h that tells us that our acceleration is zero. So from there, we have the force from our engine minus the force, The drag force is equal to zero or the force of the engine is equal to the drag force from there. We can recall that our drag force Is given by the equation 1/2 row C A V squared Moreau is the density C is our drag coefficient A is the area and V is our speed velocity. We can also recall for energy because they're asking us about the chemical energy. We can recall the energy is given as power times delta T with the change in T. And further more power is given by force times velocity. So substituting that and we get our energy is given by force time speed times delta T. So the next step is going to be calculating the engine's mechanical energy and then we will work to determine the chemical energy. So our mechanical energy is going to be equal to F. So our forces just the force of our engines that's F E times speed times delta T we've already established that are the engines force F E is equal to the drag force. And we know the equation for a drag force. So we'll use that one half rho C A B squared multiplied by V multiplied by delta T. From the problem, We were told that our density is 1.2 kg per meter cubed. Our drag coefficient was given as 0.85. Our area was given as 0. meters squared. And our speed was given at kilometers per hour. Gonna rewrite that using standard units as 50 times 10 to the third meters Per hour or 330/600, put that into our calculator and we get a speed of 13.8, 9 meters per second. Our time was given to us as 40 minutes Again, we want to get this into seconds or standard units. And so we can recall that there are 60 seconds per minute and that leaves us with 2400 seconds as our delta T okay. So from there, we can plug in everything we know back into this equation. So our mechanical energy is going to be 1/2 times 1.2 kg Per meter Cube time, 0.85 drag coefficient times our area 0.8 m squared times are speed, speed squared. Times B so it's gonna be speed cubed. So it's 13.8, m/s. Cubed times are time 24/00, Plug that into our calculator. And we are left with the mechanical energy of 2.6, 24 Times 10 to the six jewels. Now, the problem tells us that we have an efficiency of 31%. So our chemical energy Is going to be our mechanical energy divided by 0.31. And when you plug that in, that comes out to a chemical energy of 8.46 times 10 to the six jewels almost there. The last step is that they asked us to determine the chemical energy in units of kilocalories. And so we'll take 8.4, 6 times 10 to the six jewels Times a conversion factor of one killer calorie Times over 4.19 killer jewels or tend to the third jewels Put that in and we get 2. Times 10 to the 3rd killer calories. So that is the final answer for this problem. You look at our potential choices and that aligns with answer choice D. So E is the correct answer for this problem. So that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Drag Force

Work and Energy

Caloric Expenditure