Lab: Kinetic, Potential Energy and Work
Theory/Objective – Part 1
This lab is about gravitational potential energy, kinetic energy, and the relationship between them when objects change position and speed due to the effects of gravity. Potential energy is the energy of position and kinetic energy is the energy of motion. On a roller coaster potential energy depends on your height above the ground and the kinetic energy depends on how fast you are going. Both kinds of energy are measured in Joules (J). In preparation for this lab visit the website
http://adventure.howstuffworks.com/roller-coaster3.htm
and read the article “How Roller Coasters work.” Run the simulation on that same page.
Part 1 – roller coaster
Consider a roller coaster car of mass 100kg on a frictionless track at points A,B,C,D as shown below…
A
D
m=100kg
C
B
Using the speeds (v) and heights (h) given in the chart below calculate the missing values for potential energy (PE), and kinetic energy (KE) at points A,B,C,D using the equations below …Remember the mass of the car is 100kg and the acceleration due to gravity (g) is 10m/s2. Round your answers to the nearest whole number.
v(m/s) / h
(m) / PE
(J) / KE
(J)
A / 0m/s / 20m
B / 20m/s / 0m
C / 16m/s / 7.2m
D / 8m/s / 16.8m
Analysis Part 1–Complete the following calculations and questions
1. Mechanical Energy is the sum of Gravitational potential energy and Kinetic energy. In other words …
Calculate the total mechanical at each point on the coaster (A,B,C,D)
A.
B.
C.
D.
2. How does the total mechanical energy at every point on the ride compare to every other point on the coaster?
3. Consider the same roller coaster car at four new points E,F,G,H. Find the missing values at each point and complete the table. (Use g=10m/s2)
m=100kg
H
E
F
G
vm/s / h
m / PE
J / KE
J
E / 11,000J
F / 5000J
G / 0J / 20000J
H / 3000J
HINT: FIND THE MISSING ENERGY FIRST. THIS IS EASY BECAUSE THE TOTAL MECHANICAL ENERGY NEVER CHANGES. THEN USE THE EQUATIONS FOR KE AND PE TO GET THE OTHER MISSING VALUES. (remember g=10m/s2)
Part 2 – Work and the Bow and Arrow
Force(N) / 0 / 7 / 14.5 / 21 / 27.5 / 35 / 42.4 / 49 / 55.7 / 63 / 69.
Distance
(m) / 0 / 0.01 / 0.02 / 0.03 / 0.04 / 0.05 / 0.06 / 0.07 / 0.08 / 0.09 / 0.10
Hint: Be sure to include all titles, axis titles, and units
Theory – Part 2
The area of any F vs. d graph is equal to work. So the area of this graph is the work that is done in drawing back the arrow in the bow. The work that was done on the arrow has given the arrow a kind of potential energy called Elastic Potential Energy. When the arrow is released, all of this Elastic Potential energy will be converted to Kinetic Energy, the energy of motion. So…
Area of Graph = Work done = Elastic Potential energy = Kinetic Energy
In Equation form
Area (J) =
Where v is the speed of the arrow and m is the mass of the arrow which was measured to be m = 0.030kg.
Analysis – Part 2
1. Draw a best-fit line on your graph through the data. Start your line at the origin of the graph (0,0).
2. Find the area under your best-fit line on your graph in the space below. The area under your line is a triangle, and the area of a triangle is
Using the units of the axes on your graph, the area has units of JOULES.
Area - ______J
3. As described above, Area (J) = . Using this equation calculate the speed of the arrow as it exits the bow. Show your calculation in the space below.
(m=0.30kg)
v = ______m/s
4. Assume the arrow is shot horizontally from the bow and the arrow leaves the bow with the speed you calculated above. Calculate the time of flight of the arrow if it is shot from 1.5 meters off of the ground. Use a projectile motion equation and solve it for t, using the givens below.
Dy=-1.5m
viy = 0m/s
g = -9.8m/s2
t = ?
time of flight (t) = ______s
5. Now calculate the Range of the arrow. In other words, calculate how far horizontally it will fly using the projectile motion equation Dx=vxt, where vx is the speed (v) calculated in #3 and the t is the time calculated in #4. Show your work in the space below.
Dx = ______m