Created by Turbolearn AI
Work and Kinetic Energy
Calculating Work Done by a Force
To move a car of mass m = 1000 kg along a level road, a force of about P = 200 N is needed to overcome resistance. If the
resistance is R = 100 N, and the car is pushed x = 20 m:
The net force on the car is (200 − 100)N = 100N . Using Newton's Second Law: 100 = 1000a
Acceleration a: a = 100 2
= 0.1 m/s
1000
The car's final speed v: v 2
= 0 + 2ax = 2 ⋅ 0.1 ⋅ 20 = 4 v = 2 m/s
In the general case, for a car of mass m kg, with applied force P and resistance R: P − R = ma a =
P −R
m
After moving a distance x from rest, the speed v is: v 2
= 2ax = 2 ⋅
P −R
m
⋅ x (P − R)x =
1
2
mv
2
Kinetic Energy
Kinetic energy is the energy of an object due to its motion.
1 2 1 2
Kinetic Energy = ⋅ mass ⋅ (speed) Kinetic Energy = mv
2 2
The energy comes from pushing the car. Some of the work overcomes resistance, and the rest accelerates the car.
Work Done by a Constant Force
Work done by a constant force is the product of the force and the distance moved in the direction of the force.
Work = Force ⋅ distance
−Rx is the work done by the resistance. The negative sign indicates the force direction is opposite to the motion. If the work
done by a force is negative, work is done against the force.
Work-Energy Equation
The left-hand side of the equation is the work done by the resultant force in the direction of motion, and it equals:
(work done by f orce P) - (work done against f orce R) This positive work done produces the kinetic energy. In mechanics, to do
work, an object has to move through some distance.
Units
The unit of work is the joule (J). 1J is the work done in moving a force of 1N through 1m. 1J = 1N m For weights of mass 100 kg
lifted 2 m, the work done against gravity is: 100 ⋅ g ⋅ 2 = 2000J (taking g = 10 m/s ) 2
The work-energy equation relates work done and kinetic energy.
The Jumping Flea
A flea Spilopsyllus weighs about 0.4 mg and jumps at about 1 m/s. Its kinetic energy is: 0.4 mg = 0.4 ⋅ 10 kg −6
J The earthworm crawls at about 0.5 cm/s. A fast species of
1 −6 2 −6 −7
Kinetic Energy = ⋅ (0.4 ⋅ 10 ) ⋅ (1) = 0.2 ⋅ 10 = 2 ⋅ 10
2
centipede runs at 0.45 m/s.
Gravel Beds
Purpose: to bring a vehicle to a halt safely in the event of brake failure. The car enters with kinetic energy and slows down due to
the friction of the bed, losing energy. Suppose a car of mass m = 1 tonne enters the bed at v = 80 kph and stops after x = 30 m.
Kinetic energy lost: 1
2
mv
2
Page 1
, Created by Turbolearn AI
Average resistance force: R
Deceleration in terms of R and m: a = R
m
Work-energy equation: Rx = 1
2
mv
2
Green to Amber Example
A car of mass m = 1300 kg is traveling at v = 10 m/s when the lights change. The driver brakes x = 23 meters from the lights and
stops on the line.
(a) Kinetic energy before braking: ⋅ 1300 ⋅ (10)
1
2
2
= 65000J (b) Work done in bringing the car to rest: 65000J (c) Force due to
brakes: F = work
distance
=
65000
23
≈ 2826N
Forces Acting on a Car
Force Description Work Done?
Weight (mg) Force due to gravity acting downwards No
Normal Contact Force Force exerted by the surface perpendicular to the surface No
Applied Force Force applied to move the car Yes
Resistance Force opposing motion Yes
If the direction of motion is perpendicular to the force, no work is done by the force.
Car on an Icy Slope
Car mass = 1300 kg, slides 100 m down a 30 slope. ∘
Forces acting:
Weight (1300g N)
Normal contact force The normal contact force does no work. Distance moved in the direction of the weight:
100 ⋅ sin(30 ) = 50m Work done by gravity: 1300 ⋅ g ⋅ 50 = 1300 ⋅ 10 ⋅ 50 = 650000J
∘
Alternatively, consider the component of weight down the slope: 1300g sin(30 ∘
) Work done = (1300g sin(30 ∘
)) ⋅ 100 = 650000J
Summary
Work done by a force = (component of force in the direction of motion) * (distance moved).
Work Done Against Gravity: Gravitational Potential Energy
Weights of mass 100 kg are lifted 2 m: Work done against gravity = 100 ⋅ g ⋅ 2J = mass ⋅ g ⋅ vertical height
For mass m kg raised vertically through h m: Work done against gravity = mass ⋅ g ⋅ vertical height mgh is the work done against
gravity.
Car sliding down icy slope: Work done by gravity = 1300 ⋅ g ⋅ 50J = mass ⋅ g ⋅ vertical distance If the car moves up the slope:
Work done by gravity = −1300 ⋅ g ⋅ 50J = −(mass ⋅ g ⋅ vertical distance) Work done against gravity
= 1300 ⋅ g ⋅ 50J = mass ⋅ g ⋅ vertical distance
For a car of mass m, the work done against gravity is still mgh. This result is the same for any slope.
In fact, this result is true for any path joining two points A and B. The work against gravity is always
m ⋅ g ⋅ (vertical height f rom A to B), and is independent of the path joining A to B.
Energy due to position is called potential energy. Gravitational potential energy is a form of potential energy.
Gain in gravitational potential energy in moving an object from A to B: = work done against gravity in moving from A to B
= (mass) ⋅ g ⋅ (vertical height f rom A to B)
Page 2
, Created by Turbolearn AI
Work-Energy Equation. Conservation of Energy
Pushing a car: F x = 1
2
mv
2
Where F = P − R is the resultant force in the direction of motion.
If the car moves with constant acceleration, a, from a speed u to speed v over a distance of x metres: F = ma a =
F
m
2 2 2 2 F 2 2 1 2 1 2 1 2 1 2
v = u + 2ax v = u + 2 x mv = mu + 2F x mv = mu + Fx Fx = mv − mu
m 2 2 2 2
work done by the force F = final kinetic energy - initial kinetic energy work done by the force F = change in kinetic energy
produced by the force.
For a body falling under gravity from A to B, F = mg mgh =
1
2
mv
2
−
1
2
mu
2
work done by gravity = change in kinetic energy produced by gravity.
Defining mechanical energy:
Mechanical energy = kinetic energy + potential energy
Mechanical energy has the same value at A and B, meaning mechanical energy is conserved. It does not matter which horizontal
or zero you work from in calculating the potential energy since it is only changes in potential energy that are involved. Once you
have chosen the zero level, you must use the same level throughout.
When you release a ball from rest, as the height of the ball decreases its potential energy decreases, but the speed of the ball
increases, and so its kinetic energy increases. If air resistance is neglected, the only force acting is gravity. mgh = mv , u = 0 So 1
2
2
the decrease in potential energy = increase in kinetic energy. Since m cancels, v = 2gh or v = √2gh. 2
Rough Inclined Plane Example
Forces acting on the skier:
Weight (mg)
Normal reaction (N )
Friction (F ) Increase in kinetic energy = mv Distance down the slope = L =
1
2
2
N = mg cos θ F = μN = μmg cos θ
h
sin θ
Work done by weight and friction: W = (mg sin θ)L − (μmg cos θ)L W = mg sin θ − μmg cos θ
sin θ
h h
sin θ
W = mgh − μmgh cot θ By the work-energy equation: mv = mgh − μmgh cot θ v = 2gh(1 − μ cot θ) Note that v > 0
1 2 2 2
2
means 1 > μ cot θ or tan θ > μ
Power
Power is the rate at which work is done.
work done
Average power =
time taken
The unit of power is the watt (W). 1 watt is defined as 1 joule per second: 1W = 1
J
s
Power is also measured in kilowatts (kW).
Engineers often quote power in horsepower (h.p.): 1 h.p. = 746W
Log Flume Example
Time taken to raise one boat up the slope = 20 s. Angle of elevation = 30 Height = 12 m. Mass of boat = 150 kg. Average mass ∘
of each passenger = 60 kg. Energy required to raise 1 boat and 6 passengers = ? Power required to raise the full boat = ?
Power and Speed Relationship
If a constant force F moves a body with speed v: W = Fx P =
dW
dt
= F
dx
dt
= Fv P = Fv Power = Force * Speed.
Car Moving Against Resistance Example
A car moves along a horizontal road against a resistance of 400 N. Engine power = 16 kW. 16000 = 400v v = 16000
400
= 40 m/s
Maximum speed = 40 = 40 ⋅ m
= 144 kph
s
3600
1000
Page 3
Work and Kinetic Energy
Calculating Work Done by a Force
To move a car of mass m = 1000 kg along a level road, a force of about P = 200 N is needed to overcome resistance. If the
resistance is R = 100 N, and the car is pushed x = 20 m:
The net force on the car is (200 − 100)N = 100N . Using Newton's Second Law: 100 = 1000a
Acceleration a: a = 100 2
= 0.1 m/s
1000
The car's final speed v: v 2
= 0 + 2ax = 2 ⋅ 0.1 ⋅ 20 = 4 v = 2 m/s
In the general case, for a car of mass m kg, with applied force P and resistance R: P − R = ma a =
P −R
m
After moving a distance x from rest, the speed v is: v 2
= 2ax = 2 ⋅
P −R
m
⋅ x (P − R)x =
1
2
mv
2
Kinetic Energy
Kinetic energy is the energy of an object due to its motion.
1 2 1 2
Kinetic Energy = ⋅ mass ⋅ (speed) Kinetic Energy = mv
2 2
The energy comes from pushing the car. Some of the work overcomes resistance, and the rest accelerates the car.
Work Done by a Constant Force
Work done by a constant force is the product of the force and the distance moved in the direction of the force.
Work = Force ⋅ distance
−Rx is the work done by the resistance. The negative sign indicates the force direction is opposite to the motion. If the work
done by a force is negative, work is done against the force.
Work-Energy Equation
The left-hand side of the equation is the work done by the resultant force in the direction of motion, and it equals:
(work done by f orce P) - (work done against f orce R) This positive work done produces the kinetic energy. In mechanics, to do
work, an object has to move through some distance.
Units
The unit of work is the joule (J). 1J is the work done in moving a force of 1N through 1m. 1J = 1N m For weights of mass 100 kg
lifted 2 m, the work done against gravity is: 100 ⋅ g ⋅ 2 = 2000J (taking g = 10 m/s ) 2
The work-energy equation relates work done and kinetic energy.
The Jumping Flea
A flea Spilopsyllus weighs about 0.4 mg and jumps at about 1 m/s. Its kinetic energy is: 0.4 mg = 0.4 ⋅ 10 kg −6
J The earthworm crawls at about 0.5 cm/s. A fast species of
1 −6 2 −6 −7
Kinetic Energy = ⋅ (0.4 ⋅ 10 ) ⋅ (1) = 0.2 ⋅ 10 = 2 ⋅ 10
2
centipede runs at 0.45 m/s.
Gravel Beds
Purpose: to bring a vehicle to a halt safely in the event of brake failure. The car enters with kinetic energy and slows down due to
the friction of the bed, losing energy. Suppose a car of mass m = 1 tonne enters the bed at v = 80 kph and stops after x = 30 m.
Kinetic energy lost: 1
2
mv
2
Page 1
, Created by Turbolearn AI
Average resistance force: R
Deceleration in terms of R and m: a = R
m
Work-energy equation: Rx = 1
2
mv
2
Green to Amber Example
A car of mass m = 1300 kg is traveling at v = 10 m/s when the lights change. The driver brakes x = 23 meters from the lights and
stops on the line.
(a) Kinetic energy before braking: ⋅ 1300 ⋅ (10)
1
2
2
= 65000J (b) Work done in bringing the car to rest: 65000J (c) Force due to
brakes: F = work
distance
=
65000
23
≈ 2826N
Forces Acting on a Car
Force Description Work Done?
Weight (mg) Force due to gravity acting downwards No
Normal Contact Force Force exerted by the surface perpendicular to the surface No
Applied Force Force applied to move the car Yes
Resistance Force opposing motion Yes
If the direction of motion is perpendicular to the force, no work is done by the force.
Car on an Icy Slope
Car mass = 1300 kg, slides 100 m down a 30 slope. ∘
Forces acting:
Weight (1300g N)
Normal contact force The normal contact force does no work. Distance moved in the direction of the weight:
100 ⋅ sin(30 ) = 50m Work done by gravity: 1300 ⋅ g ⋅ 50 = 1300 ⋅ 10 ⋅ 50 = 650000J
∘
Alternatively, consider the component of weight down the slope: 1300g sin(30 ∘
) Work done = (1300g sin(30 ∘
)) ⋅ 100 = 650000J
Summary
Work done by a force = (component of force in the direction of motion) * (distance moved).
Work Done Against Gravity: Gravitational Potential Energy
Weights of mass 100 kg are lifted 2 m: Work done against gravity = 100 ⋅ g ⋅ 2J = mass ⋅ g ⋅ vertical height
For mass m kg raised vertically through h m: Work done against gravity = mass ⋅ g ⋅ vertical height mgh is the work done against
gravity.
Car sliding down icy slope: Work done by gravity = 1300 ⋅ g ⋅ 50J = mass ⋅ g ⋅ vertical distance If the car moves up the slope:
Work done by gravity = −1300 ⋅ g ⋅ 50J = −(mass ⋅ g ⋅ vertical distance) Work done against gravity
= 1300 ⋅ g ⋅ 50J = mass ⋅ g ⋅ vertical distance
For a car of mass m, the work done against gravity is still mgh. This result is the same for any slope.
In fact, this result is true for any path joining two points A and B. The work against gravity is always
m ⋅ g ⋅ (vertical height f rom A to B), and is independent of the path joining A to B.
Energy due to position is called potential energy. Gravitational potential energy is a form of potential energy.
Gain in gravitational potential energy in moving an object from A to B: = work done against gravity in moving from A to B
= (mass) ⋅ g ⋅ (vertical height f rom A to B)
Page 2
, Created by Turbolearn AI
Work-Energy Equation. Conservation of Energy
Pushing a car: F x = 1
2
mv
2
Where F = P − R is the resultant force in the direction of motion.
If the car moves with constant acceleration, a, from a speed u to speed v over a distance of x metres: F = ma a =
F
m
2 2 2 2 F 2 2 1 2 1 2 1 2 1 2
v = u + 2ax v = u + 2 x mv = mu + 2F x mv = mu + Fx Fx = mv − mu
m 2 2 2 2
work done by the force F = final kinetic energy - initial kinetic energy work done by the force F = change in kinetic energy
produced by the force.
For a body falling under gravity from A to B, F = mg mgh =
1
2
mv
2
−
1
2
mu
2
work done by gravity = change in kinetic energy produced by gravity.
Defining mechanical energy:
Mechanical energy = kinetic energy + potential energy
Mechanical energy has the same value at A and B, meaning mechanical energy is conserved. It does not matter which horizontal
or zero you work from in calculating the potential energy since it is only changes in potential energy that are involved. Once you
have chosen the zero level, you must use the same level throughout.
When you release a ball from rest, as the height of the ball decreases its potential energy decreases, but the speed of the ball
increases, and so its kinetic energy increases. If air resistance is neglected, the only force acting is gravity. mgh = mv , u = 0 So 1
2
2
the decrease in potential energy = increase in kinetic energy. Since m cancels, v = 2gh or v = √2gh. 2
Rough Inclined Plane Example
Forces acting on the skier:
Weight (mg)
Normal reaction (N )
Friction (F ) Increase in kinetic energy = mv Distance down the slope = L =
1
2
2
N = mg cos θ F = μN = μmg cos θ
h
sin θ
Work done by weight and friction: W = (mg sin θ)L − (μmg cos θ)L W = mg sin θ − μmg cos θ
sin θ
h h
sin θ
W = mgh − μmgh cot θ By the work-energy equation: mv = mgh − μmgh cot θ v = 2gh(1 − μ cot θ) Note that v > 0
1 2 2 2
2
means 1 > μ cot θ or tan θ > μ
Power
Power is the rate at which work is done.
work done
Average power =
time taken
The unit of power is the watt (W). 1 watt is defined as 1 joule per second: 1W = 1
J
s
Power is also measured in kilowatts (kW).
Engineers often quote power in horsepower (h.p.): 1 h.p. = 746W
Log Flume Example
Time taken to raise one boat up the slope = 20 s. Angle of elevation = 30 Height = 12 m. Mass of boat = 150 kg. Average mass ∘
of each passenger = 60 kg. Energy required to raise 1 boat and 6 passengers = ? Power required to raise the full boat = ?
Power and Speed Relationship
If a constant force F moves a body with speed v: W = Fx P =
dW
dt
= F
dx
dt
= Fv P = Fv Power = Force * Speed.
Car Moving Against Resistance Example
A car moves along a horizontal road against a resistance of 400 N. Engine power = 16 kW. 16000 = 400v v = 16000
400
= 40 m/s
Maximum speed = 40 = 40 ⋅ m
= 144 kph
s
3600
1000
Page 3
