Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Falling Ball Kinetic and Potential Energy We will use this virtual experiment to

ID: 1626587 • Letter: F

Question

Falling Ball Kinetic and Potential Energy

We will use this virtual experiment to explore the relationship between kinetic and potential energy. Kinetic and potential energy represent the forms of mechanical energy. Potential energy exists whenever an object which has mass has a position within a force field. The most everyday example of this is the position of objects in the earth's gravitational field. Kinetic Energy exists whenever an object which has mass is in motion with some velocity. Everything you see moving about has kinetic energy. If we ignore friction (forces that would cause energy to be changed from mechanical energy to another form of energy – general heat) in the system, then the total energy of the system (potential + kinetics) will remain the same. Since the total energy remains the same, yet the kinetic (or potential) energy of the system changes in value, then the other (kinetic or potential) form of energy must also change accordingly to keep the total amount of energy constant.

Instructions:

Go to http://jersey.uoregon.edu/vlab/PotentialEnergy/index.html

Set the mass 5 kg, the potential energy to 200 J and the EAS to 50%

Answer the following questions before to start the experiment

What is the height of the ball? (calculate)

How high will the ball bounce on the first bounce? (calculate)

What will be the impact velocity of the ball on the first bounce? (calculate)

Activate the step button to run the experiment and check your answers to the three questions above.

Reset the experiment and change the mass of the ball to 2 kg. How much higher did the ball rise? Will the ball have a different velocity on initial impact as in the previous case? If so, how come? If not, why not?

Explanation / Answer

(a) let the height is h

PE=mgh

h=PE/mg=4.08 m

(b) Energy lost=50 percent

Energy=100 J

mgh=100

h=100/mg=2.04 m

(c)Initial Kineic energy=100

0.5mv^2=100

we got v=6.32 m/s

if m=2 kg

height=100/(2*9.8)=5.1 m