SCP1134 Lab Manual - Lab 3 - Motion - Preparation v1(2) revamp PDF

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SCP1134 Foundations of Physics

Lab 3. Preparation

\LAB 3 PREPARATION Name: Ho KOH Student Number: 10362372 Lab Day and Time: WEDNESDAY / FRIDAY___________________ Class Cohort: RED / BLUE 1. You have a set of scales in an elevator, and you are stationary on the ground floor of a building. The scales indicate that you “weigh” 60 kg. [5 marks] (a) What are the scales actually measuring? Explain. [1.5 marks] The scales measure the normal force that it takes to support the person’s weight. (b) You stand on the scales and press the button for the 3 rd floor. Describe qualitatively what will happen during your journey from the ground floor to the third floor. You will need to describe all the (5) stages of the elevator’s motion. [1.5 marks] During the journey from the ground to the third floor, I will undergo a sense of feeling ‘heavier’, due to the acceleration increasing. Although my weight is calculated by Mass x Gravity, the scale reading changes because the persons downward force fluctuations. When inside the elevator, there are only forces acting upon my body: The Force of Gravity and the upward Force opposing the Force due to my weight. The five stages of motion are: Stage 1: Ground floor – 0 Velocity (Stationary) – 9.8 m/s2 Stage 2: Going up (Constant Acceleration) Stage 3: Midway – Constant Velocity Stage 4: Going up (Constant Deceleration) Stage 5: Reached 3rd floor – 0 Velocity (Stationary) – 9.8 m/s2

(c) If the elevator can accelerate (and decelerate) at 5.00 m/s 2, what will be the reading on the scales during all phases of your upward journey. [2 marks] In this case we will consider Newton’s Law of Motion to solve this:

Semester 2, 2017

Copyright  Dr S. Hinckley

SCP1134 Foundations of Physics

Lab 3. Preparation

During the first phase when the elevator is moving upwards starting from ground floor, the overall acceleration of me will be upward alongside with the elevator. Therefore, the ma in the formula for Newton’s 2nd Law (F=ma) is positive as it is moving upward. Since the scale is being implemented in this scenario, we have to consider the external forces which are the force of gravity acting down (-W=-mg) and the supporting normal force FS that the scale uses upward on me. Therefore, F=ma =-mg+FN. We want to know FN because this will be the number that will be read of the scale. Therefore, the formula is derived as FN = mg + ma (The scale number will be greater than the true weight) However, once the elevator reaches the 3 rd floor, the overall acceleration of me will now be downward with the elevator. Therefore, ma is negative (downward). Again, external forces acting upon me is the force of gravity which unlike the initial start is acting down (-W=-mg) and the supporting force F N that the scale applies upward on me. Therefore, from F=-ma=-mg+FN, we get the number we read of the scale (FN): FN = mg – ma (The scale number will be less than the true weight)

2. Consider the motion of the falling object in Experiment B, which is a constant acceleration problem. You need to read the lab handout to check the parameters of the falling objects motion (initial velocity, total displacement, etc). You may use the additional graphs located on the class webpage (you will need to change the scales on each axis by altering the number on each axis), or you may construct your own graphs. To help you develop your graphs, you should answer the following questions. [5 marks]

(a) What is the expected acceleration of the object (include a relevant reference)? [1.5 marks] Acceleration due to gravity = Approx. 9.8m/s2 downwards This is because the object is accelerating under the influence of gravity and therefore for every freefall objects on Earth, the acceleration will always be at a rate of 9.8 m/s2 downwards. Reference: Representing Free Fall by Graphs (n.d.). the Physics Classroom. Retrieved from http://www.physicsclassroom.com/class/1DKin/Lesson-5/Representing-Free-Fall-by-Graphs (b) How long will it take the object to fall to the ground? [1 mark] u = 0m/s a(g) = -9.8m/s/s t=? s = 1.5 m s = ut + 1/2at^2 1.5 = ½*(-9.8)t^2  t = sqrt(0.306) = 0.55 seconds (c) What will be the object’s final velocity just before it hits the floor? [1 mark] v=? u = 0 m/s a = -9.8 m/s2 t = 0.55 s v = u + at v = 0 + (-9.8)*(0.55)

Semester 2, 2017

Copyright  Dr S. Hinckley

SCP1134 Foundations of Physics

Lab 3. Preparation

v = -5.39 Therefore, the object’s final velocity is 5.39 m/s (downwards)

(d) Sketch graphs of position, velocity and acceleration vs time (one under the other to show the corresponding behaviour). You must include estimates of appropriate values of position, velocity and acceleration as a function of time. [1.5 marks]

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Signature: ______________________________________________ Date: 3/10/2017 Signature is only required for hardcopy submissions. Reference: Apparent Weight: Person on Scale in Elevator (n.d.). Retrieved from http://ecu.au.libguides.com/referencing

Semester 2, 2017

Copyright  Dr S. Hinckley...