Bryan 11 year old 6th Grade Science Presentation on Energy

otential And Kinetic Energy Example Problem - Work and Energy Examples
This entry was posted on April 30, 2017 by Todd Helmenstine (updated on September 7, 2022)
Potential energy is energy attributed to an object by virtue of its position. When the position is changed, the total energy remains unchanged but some potential energy gets converted into kinetic energy. The frictionless roller coaster is a classic potential and kinetic energy example problem.
The roller coaster problem shows how to use the conservation of energy to find the velocity or position or a cart on a frictionless track with different heights. The total energy of the cart is expressed as a sum of its gravitational potential energy and kinetic energy. This total energy remains constant across the length of the track.
Potential And Kinetic Energy Example Problem
Rollercoaster Diagram for Conservation of Energy Example Problem
Question:
A cart travels along a frictionless roller coaster track. At point A, the cart is 10 m above the ground and traveling at 2 m/s.
A) What is the velocity at point B when the cart reaches the ground?
B) What is the velocity of the cart at point C when the cart reaches a height of 3 m?
C) What is the maximum height the cart can reach before the cart stops?
Solution:
The total energy of the cart is expressed by the sum of its potential energy and its kinetic energy.
Potential energy of an object in a gravitational field is expressed by the formula
PE = mgh
where
PE is the potential energy
m is the mass of the object
g is the acceleration due to gravity = 9.8 m/s2
h is the height above the measured surface.
Kinetic energy is the energy of the object in motion. It is expressed by the formula
KE = ½mv2
where
KE is the kinetic energy
m is the mass of the object
v is the velocity of the object.
The total energy of the system is conserved at any point of the system. The total energy is the sum of the potential energy and the kinetic energy.
Total E = KE + PE
To find the velocity or position, we need to find this total energy. At point A, we know both the velocity and the position of the cart.
Total E = KE + PE
Total E = ½mv2 + mgh
Total E = ½m(2 m/s)2 + m(9.8 m/s2)(10 m)
Total E = ½m(4 m2/s2) + m(98 m2/s2)
Total E = m(2 m2/s2) + m(98 m2/s2)
Total E = m(100 m2/s2)
We can leave the mass value as it appears for now. As we complete each part, you will see what happens to this variable.
Part A:
The cart is at ground level at point B, so h = 0 m.
Total E = ½mv2 + mgh
Total E = ½mv2 + mg(0 m)
Total E = ½mv2