Energy conservation
Work is defined as the amount of energy transferred to or from an object when a force acts on it and causes it to move a certain distance in the direction of the force.
Work is measured in joules (J) or newton-meters (Nm), where one joule is equal to one newton-meter.
The formula for work is:
W = F × d × cos(θ)
where W is work
F is the force applied
d is the displacement of the object
θ is the angle between the force and displacement vectors
When the force and displacement are in the same direction, the angle between them is 0 degrees, and the cos(θ) term becomes 1, so the formula simplifies to:
W = F × d
◦ This formula is used when the force and displacement are in the same direction.
◦ In this case, work is simply the product of the force and the distance moved by the object in the direction of the force.
◦ Work can be positive, negative, or zero, depending on the direction of the force and displacement.
◦ Positive work is done when the force and displacement are in the same direction.
◦ Negative work is done when they are in opposite directions.
◦ Zero work is done when there is no displacement. ◦ The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. ◦ In other words, the total amount of energy in a closed system remains constant over time.
For example:
In a simple system consisting of a ball rolling down a hill, the ball has gravitational potential energy due to its position on the hill. As the ball rolls down the hill, this potential energy is transformed into kinetic energy as the ball gains speed.
According to the principle of conservation of energy, the total amount of energy (potential energy plus kinetic energy) in the ball remains constant throughout the process, even though the form of energy changes.
Efficiency
◦ The efficiency of a system is a measure of how well it converts energy input into useful energy output.
◦ It is defined as the ratio of the useful energy output from the system to the total energy input.
◦ The efficiency of a system is always less than 100%, since some of the energy input is lost as waste heat or other forms of energy.
◦ The formula could be represented as:
Efficiency = (useful energy output / total energy input) x 100%
Examples:
1. A light bulb has an input power of 60 watts (W) and emits 6 watts of light. What is the efficiency of the light bulb?
Solution
Useful energy output = 6 W Total energy input = 60 W
Efficiency = (useful energy output / total energy input) x 100%
Efficiency = (6 / 60) x 100%
Efficiency = 10%
2. An electric motor has an input power of 500 watts and produces 450 watts of mechanical power. What is the efficiency of the motor?
Solution
Useful energy output = 450 W Total energy input = 500 W
Efficiency = (useful energy output / total energy input) x 100%
Efficiency = (450 / 500) x 100%
Efficiency = 90%
3. A car engine has a total input power of 20,000 J and produces 8,000 J of useful work. What is the efficiency of the car engine?
Express your answer as a percentage, to the nearest whole number.
A) 25%
B) 40%
C) 60%
D) 75%
Solution
To solve this question, we need to use the formula for efficiency, which is:
Efficiency = (useful energy output / total energy input) x 100%
Plugging in the given values, we get:
Efficiency = (8,000 / 20,000) x 100%
Efficiency = 0.4 x 100%
Efficiency = 40%
Therefore, the efficiency of the car engine is 40%, which is option B.
Power
Power is the work done per unit time.
Power = work done/ time
Examples:
◦ A construction worker uses a jackhammer to break up concrete. If the jackhammer does 3,000 J of work in 30 seconds, what is its power output?
Solution
W = 3,000 J
t = 30 s
P = W/t
P = 3,000 J / 30 s
P = 100 W
Therefore, the power output of the jackhammer is 100 watts.
◦ An elevator motor lifts a 500 kg load a distance of 20 m in 10 seconds. What is the power output of the motor?
Solution
W = Fd
= m x g x d
W = 500 kg x 9.8 m/s² x 20 m
W = 98,000 J
t = 10 s
P = W/t
P = 98,000 J / 10 s
P = 9,800 W
Therefore, the power output of the elevator motor is 9,800 watts (W).
◦ A cyclist completes a 40 km race in 2 hours. If the cyclist maintains a constant speed throughout the race, what is their average power output?
Solution
d = 40 km = 40,000 m
t = 2 hours = 7,200 s
v = d/t = 40,000 m / 7,200 s
v = 5.56 m/s
W = Fd
= m x a x d
W = 70 kg x 5.56 m/s² x 40,000 m
W = 15,568,000 J
P = W/t
P = 15,568,000 J / 7,200 s
P = 2,161 W
Therefore, the average power output of the cyclist is 2,161 watts (W).
Power and velocity
The formula for work done, W, is given by:
W = Fd
where F is the force applied
d is the displacement in the direction of the force.
We can rewrite this formula using the definition of velocity,
v = d/t
d= v x t
where t is the time taken for the displacement.
This gives:
W = F x v x t
= Fvt
We can rearrange this formula to find the power output, P, which is the rate at which work is done:
P = W/t
= Fvt/t
= Fv
Therefore, the formula for power output, P, is given by.
P = Fv
Examples: 1. A weightlifter lifts a 150 kg barbell at a constant speed of 0.5 m/s. What is the power output of the weightlifter?
Solution
F = mg
= 150 kg x 9.8 m/s²
F = 1,470 N
v = 0.5 m/s
P = F x v
P = 1,470 N x 0.5 m/s
P = 735 W
Therefore, the power output of the weightlifter is 735 watts.
2. A hydraulic press exerts a force of 30,000 N on a piece of metal, compressing it by 0.05 m in 10 seconds. What is the power output of the hydraulic press?
F = 30,000 N
d = 0.05 m
t = 10 s
v = d/t
= 0.05 m / 10 s
v = 0.005 m/s
P = F x v
P = 30,000 N x 0.005 m/s
P = 150 W
Therefore, the power output of the hydraulic press is 150 watts.
Energy
We can derive the formula for gravitational potential energy,
W = F x d
For a body with mass gaining a height, the force is due to the mass. So, force= mass x gravitational field strength F=mg Replacing force with this we get, W= m x g x d As the body moves vertically upwards it gains a height. We can replace d with h. So, W= m x g x h The work done is transformed into gravitational potential energy. So, ∆EP = mg∆h
Examples: 1. A 1 kg book is lifted from the ground to a height of 2 m. What is the change in potential energy of the book?0
Solution
m = 1 kg g = 9.8 m/s² ∆h = 2 m
∆EP = mg∆h = (1 kg) (9.8 m/s²) (2 m) = 19.6 J
Therefore, the change in potential energy of the book is 19.6 J.
Kinetic energy
We know,
W = F x s F = ma Substituting,
W = m x a x s - (1)
From the third equation of motion we know,
v² = u² + 2as
Considering the body starts from rest, u = 0
v² = 2as
as = 1/2 v²
From (1)
W = m x 1/2 x v²
The total work done is the gain in kinetic energy. KE= 1/2 x m x v²
Examples: 1. A car with a mass of 1500 kg is moving at a velocity of 30 m/s. What is its kinetic energy?
Solution
KE = 1/2 × m × v²
= 1/2 × 1500 kg × (30 m/s)²
= 675,000 J Therefore, the kinetic energy of the car is 675,000 joules (J).
2. A bullet with a mass of 0.02 kg is moving at a velocity of 500 m/s. What is its kinetic energy?
Solution
KE = 1/2 × m × v²
= 1/2 × 0.02 kg × (500 m/s)²
= 2500 J Therefore, the kinetic energy of the bullet is 2500 joules (J).
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