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# Lesson 16: Kinetic and Gravitational Energy

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Whether man is disposed to yield to nature or to oppose her, he cannot do without a correct understanding of her language.

Jean Rostand

#### Gravitational Potential Energy

Gravitational potential energy (Eg or EG) is stored energy that an object has due to its position above the ground. The higher the object is above the ground, the greater the amount of gravitational potential energy that it possesses.
The mathematical equation for gravitational potential energy can be derived by starting with the equation for work.

The equation for work:

Since work and energy are two forms of the same thing:

Substituting in the equation for the force of gravity, Fg = mg, we get EG = mgh.

h is used to represent height, which is simply the vertical displacement measured in metres.

#### Example 1: Gravitational Potential Energy

Find the EG of an 8.0 kg cat that has climbed a tree to a height of 4.0 m above the ground.

Given:

m = 8.0 kg
g = 9.8 N/kg [down]
h = 4.0 m

Required:

E= ?

Solution:

EG = mgh
E= (8.0 kg)(9.8 N/kg)(4.0 m)
E= 314 J
The cat has 314 J of gravitational potential energy.

#### Example 2: Gravitational Potential Energy

Find the change in gravitational potential energy if the cat climbs another 3.0 m higher.

Given:

m = 8.0 kg
g = 9.8 N/kg [down]
Īh = 4.0 m. The Greek letter delta, Ī, means āchangeā.

Required:

ĪE= ?. In this case, we need to determine by how much the gravitational potential energy has changed. This is different from finding the total potential energy of the cat.

Solution:

ĪEG = mgĪh
ĪE= (8.0 kg)(9.8 N/kg)(3.0 m)
E= 235 J
By climbing 3.0 m higher, the cat has changed its gravitational potential energy by 235 J.

#### Practice Questions

1. How high above the ground would a 500 kg roller coaster car be if it had 61 740 J of gravitational potential energy?

Given:

Eg = 61 740 J

g = 9.8 N/kg [down]

m = 500 kg

Required:

h = ?

Solution:

h = 12.6 m

The roller coaster car is 12.6 m high.

#### Kinetic Energy

Kinetic energy is (Ek) is energy an object has due to the fact that it is moving. The faster it is moving, the more kinetic energy it has. Kinetic energy can be found using the following equation:

or

Where

m = the mass of the object in kilograms (kg)
v = the speed of the object in metres per second (m/s)

#### Example 3: Finding Kinetic Energy

How much kinetic energy does a 5.0 kg bowling ball have if it is moving at 8.0 m/s?

Given:

m = 5.0 kg
v = 8.0 m/s

Required:

Ek = ?

Solution:

Ek = 160 J
A bowling ball moving at 8.0 m/s would have 160 J of kinetic energy.

#### Example 4: Finding Speed

A 1200 kg car is being driven down a road. If it has 101 kJ of kinetic energy, what is its speed?

Given:

m = 1200 kg
Ek = 101 kJ = 101 000 J

Required:

v = ?

Solution:

Rearrange the equation to get

v = 13.0 m/s

The car’s speed, as it is driven down the road, is 13.0 m/s.

#### Practice Questions

1. A baseball pitcher throws a baseball, which has a mass of 0.145 kg, at a speed of 153 km/h. What is the kinetic energy of the baseball?

Given:

m = 0.145 kg

v = 153 km/h = 42.5 m/s

Required:

Ek = ?

Solution:

Ek = 131 J

A baseball thrown at 153 km/h has 131 J of kinetic energy.

2. How fast would a shot putter have to throw a shot put, which has a mass of 7.26 kg, so that it would have the same amount of kinetic energy as the baseball in the above question?

Given:

Ek = 131 J (from previous question)

m = 7.26 kg

Required:

v = ?

Solution:

v = 6.0 m/s

A shot put would have to be moving at 6.0 m/s to have the same amount of kinetic energy as a baseball moving at 153 km/h, or 42.5 m/s.

#### Energy Transformations

When the girl on the toboggan was at the top of the hill, she had a great deal of gravitational potential energy due to her height. As she starts sliding down the hill, she loses gravitational potential energy, because her height is being reduced, but gains kinetic energy, because she is moving faster. You can use your knowledge of the conservation of energy and your understanding of how energy is transformed from one form into another to make predictions about her motion.

#### Practice Questions

1. Sheila and her toboggan have a total mass of 45 kg. At one point during her toboggan ride down a frictionless hill, she is 10.0 m higher than the bottom of the hill and moving at a speed of 8.0 m/s.a) How much gravitational potential energy does she have?
b) How much kinetic energy does she have?
c) How much energy does she have in total?
d) How fast will Sheila be moving when she reaches the bottom of the hill?
e) How high is the hill?

Given:

m = 45 kg

h = 10.0 m

v = 8.0 m/s

g = 9.8 N/kg [down]

Required:

1. Eg = ?
2. E= ?
3. ET =?

Solution:

1. Eg = mgh

Eg = (45 kg)(9.8 N/kg)(10.0 m)

Eg = 4 410 J

E= 1 440 J

1. ET = Eg +Ek

ET =4 410 J + 1 440 J

ET = 5 850 J

Required:

1. v = ? By the time Sheila reaches the bottom of the hill, all of her energy will be in the form of kinetic energy.

Solution:

EK(at bottom of hill) = ET = 5 850 J

v =16.1 m/s

Required:

1. h = ? At the top of the hill, all of Sheila’s energy will be in the form of gravitational potential energy.

Solution:

Eg(at top of hill) = ET = 5 850 J

h = 13.3 m

Sheila will be travelling at a speed of 16.1 m/s by the time she reaches the bottom of the hill. The height of the hill is 13.3 m.