Law of Conservation of Energy

CSEC Physics Syllabus - Effective for examinations from May - June 2015

Section A - Mechanics

Energy

Potential Energy, Ep

Specific Objective 5.6

define potential energy;

Specific Objective 5.7

calculate the change in gravitational potential energy;

Kinetic Energy, Ek

Specific Objective 5.8

define kinetic energy;

Specific Objective 5.9

calculate kinetic energies;

Conservation

Specific Objective 5.10

apply the law of conservation of energy;

Power, P

Specific Objective 5.11

define power and apply definition;

Specific Objective 5.12

explain the term efficiency;

Specific Objective 5.13

calculate efficiency in given situations.

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Example 1

An object of mass 4.0 kg is dropped from a height of 1 000 m from a helicopter. The air resistance acting against the parachute exerts a steady upward force of 35 N. Calculate the kinetic energy of the object when it reaches the ground and hence find its velocity. 

Use g = 10 N/kg

    Solution

The Law of Conservation of Energy states that energy can not be created nor destroyed but rather changes from one form to the next.

Using this law,

Work Done against Gravity = Kinetic Energy of object

F × d = K.E.

Resultant Force × d = K.E.

Resolving the Forces vertically,

K.E. = (Weight of object - 35 N) × 1 000 m

K.E. = ([m × g] - 35 N) × 1 000 m

K.E. = ([4.0 kg × 10 N/kg] - 35 N) × 1 000 m

K.E. = (40 N - 35 N) × 1 000 m

K.E. = 5N × 1 000 m

⸫ K.E. = 5 000 J

Recall,

K.E. = 1/2 × m × v²

5 000 J = 1/2 × 4.0 kg × v²

v² = 5 000 J ÷ (1/2 × 4.0 kg) 

v² = 2 500 J/kg

v = √2 500 J/kg

⸫ v = 50 m/s

Example 2

How much energy would your body need to provide for the following:

a)    climbing a hill 500 m high (use your own mass and g = 10 N/kg)

b)    sweating and evaporation of 1 kg of water from your skin

c)    heat loss by radiation at the rate of 100 watts for 12 hours. 

       (The specific latent heat of vaporization of water is 2.3 MJ/kg)

    Solution

a)    Climbing a hill 500 m high would result in a gain in gravitational potential energy

G.P.E. = m × g × h

Let us assume my mass is 82.0 kg,

G.P.E. = 82.0 kg × 10 N/kg × 500 m

⸫ G.P.E. = 410 000 J

b)    Sweating and evaporation of 1 kg of water from your skin

Energy = mass × specific latent heat of vaporization

Energy = 1 kg × 2.3 MJ/kg

⸫ Energy = 2.3 MJ

c)    Heat loss by radiation at the rate of 100 watts for 12 hours

Converting 12 hours into seconds,

1 hour = 60 minutes

1 minute = 60 seconds

⸫ 1 hour = 60 × 60 seconds

1 hour = 3 600 seconds

12 hours = 12 × 3 600 seconds

⸫ 12 hours = 43 200 seconds

Power is defined at the rate at which work is done or the energy used in a specified time.

⸫ Power = Energy ÷ Time 

Energy = Power × Time 

Energy = 100 W × 43 200 s

⸫ Energy = 4.32 MJ

Example 3

State the energy changes that occur when:

a)    gas is used to boil water in a whistling kettle

b)    a child blows up a balloon and then releases it, letting it fly around the room

c)    a guitarist plays an electric guitar over a public address system.

    Solution

a)    Pressurized gas in the tank has chemical potential energy, which is then transformed into heat energy when it is ignited. This heat energy is transferred to the kettle by the process of radiation. The heat energy absorbed by the kettle is transferred to the water in the kettle by the process of conduction. The water's thermal energy increases and as a result its temperature increases to above 100 degrees celsius where the water will undergo a phase change from liquid to vapor. This vapor has thermal and kinetic energy which rushes out the kettle and as a result, some of the kinetic and thermal energy of the vapor is transformed into sound energy as it exits the kettle. 

b)    The child blows air, which has thermal and kinetic energy into the balloon. As air is continually blown into the balloon its volume and the air pressure in the balloon increases. When the child releases the balloon, it flies around the room because the air exits using its kinetic energy where some of it is transformed into sound energy.

c)   The guitarist uses his mechanical energy to play the electric guitar. The mechanical energy transferred to the strings of the guitar is transformed into kinetic energy when they vibrate. These vibrations are then transformed into electrical energy by the electric guitar and the electrical signals are sent to the public address system. The electrical energy in the public address system is then transformed into sound and heat energy to be broadcasted by the speakers. 

Example 4

Water, which flows at the top of a waterfall at a rate of 800 kg/s, takes 1.2 s to fall vertically into the stream below.

a)    State the energy transformations that occur as the water flows and falls from the top of the waterfall into the stream below.

b)    1.    At what speed does the falling water hit the stream?

       2.    What is the height through which the water falls?

c)    Calculate:

        1.    the weight of water falling over the waterfall in 1 minute

        2.    the potential energy lost by this water when it reaches the stream below the waterfall

        3.    the power of the falling water at the instant it hits the stream.

                (Take the acceleration of free fall to be 10 m/s².)

    Solution

a)    Kinetic Energy as it flows along the river atop the waterfall -> Gravitational Potential Energy as it begins to fall off the edge of the waterfall -> Transformation from Gravitational Potential Energy into Kinetic Energy as the water falls the height of the waterfall -> All the Gravitational Potential Energy is transformed into Kinetic Energy when it hits the stream below. 

b)    1.    At the point time = 0, when the water is about to fall down the waterfall, the initial downward velocity of the water = 0. This is because the water's velocity is in the direction of the flow of the river and not down the waterfall. Gravity acting on the water causes it to gain velocity as it falls down the waterfall. 

Final Velocity = Initial Velocity + (Acceleration × Time Taken)

v = u + at

v = 0 + (10 m/s² × 1.2 s)

⸫ v = 12 m/s

        2.    Using the Law of Conservation of Energy 

Loss in Gravitational Potential Energy of the water = Gain in Kinetic Energy of the water

m × g × h =  1/2 × m × v²

We have mass, m on both sides of the equation; therefore, it could be simplified into:

g × h =  1/2 × v²

10 m/s² × h =  1/2 × (12 m/s)²

h = [1/2 × (12 m/s)²] ÷ 10 m/s²

⸫ h = 7.2 m

c)    1.    Weight of the water falling in 1 minute

1 minute = 60 seconds

Mass of water flowing in 1 minute, 

Mass = Flow rate × Time 

Mass = 800 kg/s × 60 s

⸫ Mass = 48 000 kg

        2.    The potential energy lost by this water when it reaches the stream below the waterfall

Loss in Gravitational Potential Energy = m × g × h

Loss in Gravitational Potential Energy = 48 000 kg × 10 m/s² × 7.2 m

⸫ Loss in Gravitational Potential Energy = 3.456 MJ

        3.    The Power of the falling water at the instant it hits the stream

Power = Energy ÷ Time

Power = 3.456 MJ ÷ 1.2 s

⸫ Power = 2.88 MW