Assignment 1. Problem 1: Work A 1000 kg elevator accelerates upwa...

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Assignment 1. Problem 1: Work A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. (a) Draw a before and after energy diagram. Include all known and unknown quantities. (b) Draw a free body diagram for the elevator. Don’t forget your axes. (c) How much work does gravity do on the elevator? (d) How much work does the tension in the elevator cable do on the elevator? (e) Sense making: Check your signs. Why are the signs for the both works you calculated, positive or negative? How do you know? (f) What is the elevators kinetic energy after traveling 10 m? 2. Problem 2: Graphing Work A particle moving on the x-axis experiences a force given by Fx = qx2 , where q is a constant. (a) Sketch a graph of the force exerted on the particle. (b) How much work is done on the particle as it moves from x = 0 to x = d? 3. Problem 3: Power A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. (a) What is the maximum acceleration of the Porsche on a concrete surface where µs = 1.00? Hint: What force pushes the car forward? (b) If the Porsche accelerates at amax, what is its speed when it reaches maximum power output? (c) How long does it take the Porsche to reach the maximum power output? (d) Sense making: Find your cars speed from the time you found in part (c) and the acceleration you found in part (a). Does this speed seem reasonable? ………………………………………………............................................................................................................ Page 2 of 6 Solution 1. Problem 1: Work A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. ……………………………………………….......................................................................................................... (a) Draw a before and after energy diagram. Include all known and unknown quantities . 𝕎𝑔: Gravity Force Work , 𝕎𝑇: Tension Force Work , 𝐾: Kinetic Energy ……………………………………………….......................................................................................................... (b) Draw a free body diagram for the elevator. Don’t forget your axes. Free Body Diagram (FBD) Weight: ⃗𝑾 Tension: ⃗𝑻 Displacement: ∆𝒓⃗⃗⃗⃗ ………………………………………………............................................................................................................ (c) How much work does gravity do on the elevator? 𝑊⃗⃗ 𝑻⃗ ∆⃗⃗⃗⃗𝒓 x y elevator x 𝕎 𝑎𝑛𝑑 𝐾 𝕎𝑔 𝕎𝑔 𝕎𝑇 𝕎𝑇 𝐾 𝐾 after x= 10 m before x= 10 m Page 3 of 6 Applying the definition of Work (𝕎) of a force: Work does gravity do on the elevator: 𝕎𝑔 = 𝑊⃗⃗ ∙ ∆⃗⃗⃗⃗𝑟 → 𝕎𝑔 = |𝑊⃗⃗ ||∆⃗⃗⃗⃗𝑟 |𝑐𝑜𝑠180° → 𝕎𝑔 = −𝑚𝑔|∆⃗⃗⃗⃗𝑟 | , where 𝑚 = 1 000 𝑘𝑔 is the elevator mass, 𝑔 = 9.8067 𝑚 𝑠 2 ⁄ is the standard gravity, and |∆⃗⃗⃗⃗𝑟 | = 10 𝑚 is the displacement magnitude. Replacing the numerical values, we get: 𝕎𝑔 = −1 000(9.8067)10 → 𝕎𝒈 = −𝟗𝟖 𝟎𝟔𝟕 𝑱 , rounding to 2 significant figures ( 2 sfg) , since it is the minimum number of sfg in the data, we write: 𝕎𝒈 = −𝟗𝟖 𝟎𝟎𝟎 𝑱 ………………………………………………............................................................................................................ (d) How much work does the tension in the elevator cable do on the elevator? Similarly to question above: Work the tension in the elevator cable does on the elevator: 𝕎𝑇 = 𝑇⃗ ∙ ∆⃗⃗⃗⃗𝑟 → 𝕎𝑇 = |𝑇⃗ ||∆⃗⃗⃗⃗𝑟 |𝑐𝑜𝑠0° → 𝕎𝑇 = 𝑇|∆⃗⃗⃗⃗𝑟 | To find the value of 𝑇 , from Newton’s second law, we get [see FBD in Part (b)]: 𝑇 − 𝑊 = 𝑚𝑎 , where 𝑎 is the elevator acceleration → 𝑇 = 𝑚(𝑔 + 𝑎) , then replacing this in the equation for 𝕎𝑇 . we obtain: 𝕎𝑇 = 𝑚(𝑔 + 𝑎)|∆⃗⃗⃗⃗𝑟 | which, under replacement of the numerical values, becomes: 𝕎𝑇 = 1 000(9.8067 + 1.0)10 → 𝕎𝑇 = 108 067 𝐽 , and rounding to 2sfg, we have: 𝕎𝑻 = 𝟏𝟏𝟎 𝟎𝟎𝟎 𝑱 ………………………………………………............................................................................................................ (e) Sense making: Check your signs. Why are the signs for the both works you calculated, positive or negative? How do you know? The signs for 𝕎𝒈 < 0 is correct since the angle between 𝑊⃗⃗ and ∆⃗⃗⃗⃗𝑟 is 180° [see FBD in Part (b)] , hence applying the definition of scalar product for the two vectors, 𝑐𝑜𝑠180° = −1 For 𝕎𝑇 > 0 , the reasoning is similar, only that in this case, the angle between 𝑇⃗ and ∆⃗⃗⃗⃗𝑟 is 0°, and 𝑐𝑜𝑠0° = 1 ………………………………………………............................................................................................................ (f) What is the elevators kinetic energy after traveling 10 m? Applying the Work-Energy Theorem: 𝕎𝑛𝑒𝑡 = ∆𝐾 , where 𝕎𝑛𝑒𝑡 is the Net work (sum of the works of all of the acting forces) and ∆𝐾 is the change in Kinetic Energy: 𝐾 = 𝑚𝑣 2⁄2 , being 𝑣 the speed of the body. Therefore, 𝕎𝑛𝑒𝑡 = 𝐾10 − 𝐾0 where 𝐾10 is the elevators Kinetic energy after traveling 10 m, and 𝐾0 = 1 2 𝑚𝑣0 2 = 0 , since the initial speed is 𝑣0 = 0 , hence: Page 4 of 6 𝕎𝑛𝑒𝑡 = 𝕎𝑔 + 𝕎𝑇 = 𝐾10 Using the numerical values from above, we find: 𝕎𝑛𝑒𝑡 = −98 067 𝐽 + 108 067 = 𝐾10 → 𝑲𝟏𝟎 = 𝟐 𝟎𝟎𝟎 𝑱 ………………………………………………............................................................................................................ 2. Problem 2: Graphing Work A particle moving on the x-axis experiences a force given by Fx = qx2 , where q is a constant. ………………………………………………............................................................................................................ (a) Sketch a graph of the force exerted on the particle. The Graph 𝐹(𝑥) = 𝑞𝑥 2 is a parabola. ………………………………………………............................................................................................................ (b) How much work is done on the particle as it moves from x = 0 to x = d? For a variable Force like this one: 𝐹(𝑥) = 𝑞𝑥 2 , its Work ( 𝕎𝐹 ) is calculated through the integral: 𝕎𝐹 = ∫ 𝐹(𝑥)𝑑𝑥 𝑥=𝑑 𝑥=0 → 𝕎𝐹 = ∫ 𝑞𝑥 2𝑑𝑥 𝑥=𝑑 𝑥=0 → 𝕎𝐹 = 𝑞 𝑥 3 3 | 𝑥=0 𝑥=𝑑 → 𝕎𝑭 = 𝟏 𝟑 𝒒𝒅 𝟑 ………………………………………………............................................................................................................ 3. Problem 3: Power A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. ………………………………………………............................................................................................................ (a) What is the maximum acceleration of the Porsche on a concrete surface where µs = 1.00? Hint: What force pushes the car forward? Page 5 of 6 In the figure is the FBD, where 𝑊⃗⃗ = 𝑀𝑣𝑔 , 𝑀𝑣 is the vehicle mass, 𝑔 is the gravity acceleration with magnitude 𝑔 = 9.8067 𝑚 𝑠 2 ⁄ (the standard gravity), 𝐹 𝑠𝑓𝑑 is the Static friction force and 𝑁⃗ 𝑑𝑤 is the normal force, both acting on the driving wheel, and 𝐹 𝑠𝑓𝑠 is the Static friction force and 𝑁⃗ 𝑠𝑤 is the normal force, both acting on the steering wheel. We assume that the friction and normal forces acting on each pair of wheels are represented by one force, as seen in the figure. It might appear strange that, in this model, a pushing Force, apart from the static Friction, does not appear in the FBD. To explain this, let us remember that Frictions forces are reactive forces, in the sense that the Torque that exerts the motor on the driving wheels makes a Friction force to appear, which is the one that impulses the car forward. Think for a moment what would happen with the motor working full power and no Friction at all: there would not be motion of the vehicle at all. The maximum vehicle acceleration is obtained when the Static friction force is the maximum on both kind of wheels, the driving and the steering. Their magnitudes are given by: |𝐹 𝑠𝑓𝑑| = 𝐹𝑠𝑓𝑑 = 𝜇𝑠 |𝑁⃗ 𝑑𝑤| = 𝜇𝑠𝑁𝑑𝑤 , |𝐹 𝑠𝑓𝑠| = 𝐹𝑠𝑓𝑠 = 𝜇𝑠 |𝑁⃗ 𝑠𝑤| = 𝜇𝑠𝑁𝑠𝑤 [Eqs 3a-1] Applying Newton’s second law, we have: x-axis: 𝐹𝑠𝑓𝑑 − 𝐹𝑠𝑓𝑠 = 𝑀𝑣𝑎 , y-axis: 𝑊 − 𝑁𝑑𝑤 − 𝑁𝑠𝑤 = 0 [Eqs 3a-2] Substituting the Eqs 3a-1 in the Eqs 3a-2, we find: 𝐹𝑠𝑓𝑑 − 𝐹𝑠𝑓𝑠 = 𝑀𝑣𝑎 → 𝜇𝑠𝑁𝑑𝑤 − 𝜇𝑠𝑁𝑠𝑤 = 𝑀𝑣𝑎 → 𝜇𝑠 (𝑁𝑑𝑤 − 𝑁𝑠𝑤) = 𝑀𝑣𝑎 [Eq 3a-3] 𝑊 − 𝑁𝑑𝑤 − 𝑁𝑠𝑤 = 0 → 𝑀𝑣𝑔 − 𝑁𝑑𝑤 − 𝑁𝑠𝑤 = 0 [Eq 3a-4] From Eq 3a-4 , we get 𝑁𝑠𝑤 = 𝑀𝑣𝑔 − 𝑁𝑑𝑤 , that substituted in Eq 3a-3 gives: 𝜇𝑠 [𝑁𝑑𝑤 − (𝑀𝑣𝑔 − 𝑁𝑑𝑤)] = 𝑀𝑣𝑎 → 𝜇𝑠 (2𝑁𝑑𝑤 − 𝑀𝑣𝑔) = 𝑀𝑣𝑎 From which: 𝑎 = 𝜇𝑠 (2 𝑁𝑑𝑤 𝑀𝑣 − 𝑔) According to the problem “ . . . two-thirds of the weight is over the drive wheels” , then 𝑁𝑑𝑤 = 2 3 𝑊 = 2 3 𝑀𝑣𝑔 which, replaced in former equation produces: 𝑎 = 𝜇𝑠 [2 ( 2 3 𝑔) − 𝑔] → 𝒂 = 𝟏 𝟑 𝝁𝒔𝒈 𝑭⃗ 𝒔𝒇𝒅 𝑵⃗⃗ 𝒅𝒘 driving wheels car steering wheels 𝑭⃗ 𝒔𝒇𝒔 𝑵⃗⃗ 𝒔𝒘 𝑊⃗⃗ x y Page 6 of 6 Replacing the data: 𝜇𝑠 = 1.00 and 𝑔 = 9.8067 𝑚 𝑠 2 ⁄ , we find: 𝑎 = 1 3 1.00(9.8067) → 𝑎 = 3.2689 𝑚 𝑠 2 Rounding to 3 sfg, we have: 𝒂 = 𝟑. 𝟐𝟕 𝒎 𝒔 𝟐 ………………………………………………............................................................................................................ (b) If the Porsche accelerates at amax, what is its speed when it reaches maximum power output? The effective Power out-put ( 𝒫𝑒𝑓𝑓) , the one that reaches the wheels, more specifically the driving wheels, is given by: 𝒫𝑒𝑓𝑓 = 𝜂𝒫𝑜𝑢𝑡 where the Efficiency is 𝜂 = 70% , it is 𝜂 = 0.70 , and 𝒫𝑜𝑢𝑡 = 217 ℎ𝑝 = 161 820 𝑊 ( 1 ℎ𝑝 = 745.70 𝑊) The value for the Efficiency comes from: “ . . . of the power . . . 70% reaches the wheels” following the Problem redaction. Now, 𝒫𝑒𝑓𝑓 = 𝐹 𝑒𝑓𝑓 ∙ 𝑣 , where 𝐹 𝑒𝑓𝑓 is the Effective force that pushes the car forward and 𝑣 is the car velocity. Both vectors are parallel then, 𝒫𝑒𝑓𝑓 = 𝐹𝑒𝑓𝑓𝑣, with |𝐹 𝑒𝑓𝑓| = 𝐹𝑒𝑓𝑓 and |𝑣 | = 𝑣 and using 𝒫𝑒𝑓𝑓 = 𝜂𝒫𝑜𝑢𝑡 , we get: 𝜂𝒫𝑜𝑢𝑡 = 𝐹𝑒𝑓𝑓𝑣. From Newton’s second law, 𝐹𝑒𝑓𝑓 = 𝑀𝑣𝑎 , therefore 𝜂𝒫𝑜𝑢𝑡 = 𝐹𝑒𝑓𝑓𝑣 = 𝑀𝑣𝑎𝑣 , to finally obtain 𝒗 = 𝜼𝓟𝒐𝒖𝒕 𝑴𝒗𝒂 With the numerical data, we have: 𝑣 = 0.70(161 820) 1 480(3.2689) → 𝑣 = 23.414 𝑚 𝑠 ,which rounded to 3 sfg is: 𝒗 = 𝟐𝟑. 𝟒 𝒎 𝒔 ………………………………………………............................................................................................................ (c) How long does it take the Porsche to reach the maximum power output? The time it takes to the vehicle to reach the this maximum power output is the same that employs to reach the speed just calculated, departing from rest (initial speed 𝑣0 = 0) . From the Kinematic equation: = 𝑣0 + 𝑎𝑡 , we obtain 𝑡 = 𝑣⁄𝑎 , and replacing the already used data, we find: 𝑡 = 23.414 3.2689 → 𝑡 = 7.2617 𝑠 that rounded to 3 sfg is 𝒕 = 𝟕. 𝟐𝟔 𝒔 ………………………………………………............................................................................................................ (d) Sense making: Find your cars speed from the time you found in part (c) and the acceleration you found in part (a). Does this speed seem reasonable? The car speed obtained in Part (b) was obtained with the acceleration from Part (a), and the time found in Part (c) was gotten with the speed which is suggested to use, then we are corroborating the value of the speed through the time that has just been obtained. The speed from Part (b) 𝑣 = 23.414 𝑚⁄𝑠 corresponds to 84.3 𝑘𝑚⁄ℎ , it is to say that this car goes from zero to 84.3 𝑘𝑚⁄ℎ in just 7.26 s, which is perfectly reasonable for a car with such a power out-put.

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