Potential And Kinetic Energy Example Problem – Work and Energy Examples


Potential energy is energy attributed to an object by virtue of its position. When the position is changed, the total energy remains unchanged but is converted to a different type of energy, like 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 ProblemQuestion:

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 

All of the energy at this point is kinetic energy. Since total energy is conserved, the total energy at point B is the same as the total energy at point A.

Total E at A = Total Energy at B
m(100 m2/s2) = ½mv2

Divide both sides by m
100 m2/s2 = ½v2

Multiply both sides by 2
200 m2/s2 = v2

v = 14.1 m/s

The velocity at point B is 14.1 m/s.

Part B:

At point C, we know only a value for h (h = 3 m).

Total E = ½mv2 + mgh
Total E = ½mv2 + mg(3 m)

As before, the total energy is conserved. Total energy at A = total energy at C.

m(100 m2/s2) = ½mv2 + m(9.8 m/s2)(3 m)
m(100 m2/s2) = ½mv2 + m(29.4 m2/s2)

Divide both sides by m

100 m2/s2 = ½v2 + 29.4 m2/s2
½v2 = (100 – 29.4) m2/s2
½v2 = 70.6 m2/s2
v2 = 141.2 m2/s2
v = 11.9 m/s

The velocity at point C is 11.9 m/s.

Part C:

The cart will reach its maximum height when the cart stops or v = 0 m/s.

Total E = ½mv2 + mgh
Total E = ½m(0 m/s)2 + mgh
Total E = mgh

Since total energy is conserved, the total energy at point A is the same as the total energy at point D.

m(100 m2/s2) = mgh

Divide both sides by m

100 m2/s2 = gh

100 m2/s2 = (9.8 m/s2) h

h = 10.2 m

The maximum height of the cart is 10.2 m.

Answers:

A) The velocity of the cart at ground level is 14.1 m/s.
B) The velocity of the cart at a height of 3 m is 11.9 m/s.
C) The maximum height of the cart is 10.2 m.

This type of problem has one main key point: total energy is conserved at all points of the system. If you know the total energy at one point, you know the total energy at all points.

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