Lab-5_-WORK ENERGY THEOREM

docx

School

University of Texas, Arlington *

*We aren’t endorsed by this school

Course

1443

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

docx

Pages

Uploaded by mannanzahid203

Lab-05 Work-Energy Theorem Work Energy Theorem Objective The objective of this experiment is to examine the conversion of work into kinetic energy, specifically work done by the force of gravity. The work-kinetic energy theorem equates the net force (gravity, friction, air resistance, etc.) acting on a particle with the kinetic energy gained or lost by that particle. KE = 1 2 m v 2 ( 1 ) PE g = mgh ( 2 ) If there are no other external forces acting on the system and doing work, the sum of these two forms of energy is conserved. Theory If an amount of work, W, is done on a system, then the kinetic energy, KE, of the system changes. The change in the kinetic energy is equal to the amount of work done. Stated quantitatively: W = ∆ KE = 1 2 mv f 2 − 1 2 mv i 2 ( 3 ) Where, v f and v i are the final and initial velocities of the object. Equation 3, is a statement of the work-energy theorem and is the foundation of much of our science. As such, questions of its validity are certainly worth your consideration. The Work-Energy Theorem presents a way of dealing with kinematic quantities in mechanics without regard for vector direction. These directionless quantities, such as kinetic energy, are called scalars. In this lab, we will test the work energy theorem using an object sliding the ramp (Figure 1). When there is no friction, the object will slide due to the gravitational force component acting along the surface ( mg sin θ ). Knowing the distance d traveled along the surface, we can calculate the work: W = Fd = ¿ (4)

WORK ENERGY THEOREM Page 2 The object will accelerate due to the acceleration vector component along the surface, which can be calculated using a x = g sin θ (5) Thus, the object will accelerate from the rest and have a velocity v f = v i + a x t (6) Therefore, if we know the time it takes the object to reach the Figure 1 bottom of the ramp, we can calculate the final velocity. Lab Report Section: ____005_________ Name: __________Muhammad Zahid________ GOAL: (briefly state what experiment(s) will be performed and with what purpose) Procedure: Go to: https://phet.colorado.edu/sims/cheerpj/the-ramp/latest/the-ramp.html?simulation=the-ramp 1. Note that gravity in this lab is 9.8 m/s 2 2. Select the filing cabinet for the object on the ramp. 3. Leave the position equal to 10 m (bear in mind that this is not the top of the ramp). 4. Change the angle to 10 0 . Part A: 5. Calculate the height above the ground at position 10 m: ___1.74____. 6. Minimize the force graph and open the energy graph. 7. The simulation will run when you select “frictionless” . Pause it just before the filing cabinet reaches the bottom of the ramp. 8. Read off the simulation graph the following energy values: Kinetic Energy (Top): ______0__________________ Potential Energy (Top): _____1701.75__________________

WORK ENERGY THEOREM Page 3 Total Mechanical Energy (Top): ____1701.75____________ 9. Based on the above data (see question 8), what should the energy values be at the bottom of the ramp: Kinetic Energy (Bottom): _______1684.71_________________ Potential Energy (Bottom): _______0________________ Total Mechanical Energy (Bottom): _______1684.71_________ 10. Using the height you calculated in question 5, calculate the energy values at the top and at the bottom of the ramp. Explain why there are differences between the graph values and the answers you received. Top ke = 0 J pe = 1705.75 J Bottom ke = 1693.21 J pe = 0 J It’s hard to use the simulator to get the exact values which is why there’s a difference in the graph values and answers I received. 11. Calculate the speed of the filing cabinet at the bottom of the ramp, using equation 3 from the theory. 5.8 m/s 12. Use the distance the cabinet traveled down the ramp, its initial speed down the ramp, and the time it took the cabinet to go down the ramp (found in the top left corner) to also calculate final speed (use equations 5 and 6 from theory). 5.88 m/s 13. Compare the final speeds you got from #11 and #12. Explain the difference. They are the same speeds. Part B 14. Reset the simulation. Minimize the energy graph and open the work graph. 15. Read off the chart before you run the simulation: Work done by gravity = _____-1701.75 J________ 16. Show below how you can calculate this value:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

WORK ENERGY THEOREM Page 4 Mg x 10sin(10) = 1701.75 17. If the work done by gravity already has a value for the filing cabinet sitting at rest on the ramp, what can we assume happened to give it this value? (We would think it would be zero before we start the simulation) The energy is sitting there waiting for a force to turn into kinetic energy. 18. Calculate the weight of the filing cabinet parallel to the ramp. 170.18 N 19. What distance would you multiply with the weight parallel (#18 answer) to get the work done by the weight? 10 20. Calculate below the work done by the weight. 1701.8 21. Click frictionless to run the simulation and pause it when the cabinet reaches the bottom of the ramp. 22. Read off the graph: work done by gravity = ____-0.00________ How did the work done by gravity become this value? Turned into potential energy so work is 0 23. How does the work done by gravity on the box compare to the energies of the box (think about the formulas we know)? __equal to the gravitational pe of the box_____________________________________ 24. Write down your General Observations and Conclusion.

WORK ENERGY THEOREM Page 5 We can use the work-energy theorem to calculate ke with displacement. As an object moves down a ramp, pe turns into ke. The change in the ke is the amount of work that is done. An objects weight, distance traveled aka displacement, and acceleration can all affect amount of work done. Application Questions: Assume you have a cart with a mass m = 0.530 kg. The cart accelerates from rest to a final velocity of 2.1 m/s. 1. What is the net work done on the cart? 1.169 J 2. If the work is done over the displacement of 0.71 m, what is the average net force on the cart? 1.646 N 3. If there is an uncompensated frictional force f k = 0.11 N, what is the work done by friction over the same displacement? .078 J

WORK ENERGY THEOREM Page 6 Once you have completed the assignment, save your report as a PDF and upload using the blackboard submission on the lab course page (found under “Lab Online Submissions”)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version