Work and Energy Notes
Energy
Energy Definition:
The capacity to perform work, which involves applying a force over a distance causing displacement.
Storable and transformable, existing in various forms (e.g., potential, kinetic, thermal, electrical, chemical, nuclear).
SI unit: Joule (J), defined as the work done when a force of one newton displaces an object by one meter (1 J = 1 N \cdot m).
Mechanical Energy Definition:
The energy of an object due to its motion (kinetic energy) or position (potential energy).
Sum of kinetic and potential energy in a system, representing the total energy available for mechanical work.
Types of Mechanical Energy
Gravitational Potential Energy:
Energy stored in an object due to its height above a reference point, representing the potential to do work by falling.
PE = mgh
m = mass (kg)
g = gravity (9.8 m/s² on Earth, representing the acceleration due to gravity)
h = height (m) above the reference point
Kinetic Energy:
Energy possessed by an object due to its motion, defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
KE = \frac{1}{2}mv^2
m = mass (kg)
v = velocity (m/s), indicating the speed and direction of motion
Example of Energy in Sports
Bicyclists using gravitational potential energy by gaining height on a ramp and then converting it to kinetic energy to perform tricks without pedaling.
Various sports harnessing gravity's power, such as skiing, diving, and basketball, where potential energy is converted into kinetic energy during descent or motion.
Law of Conservation of Energy
Energy Conservation states that energy cannot be created or destroyed, only transformed from one form to another within an isolated system.
Total energy in a closed system remains constant, although it may change forms, which is fundamental in physics.
MEi = MEf
KEi + PEi = KEf + PEf
The total mechanical energy of a system remains constant if only conservative forces (like gravity, spring force) are present, and non-conservative forces (like friction, air resistance) are negligible.
Energy Transformation
Skateboarder in a halfpipe: Potential energy at the top converts to kinetic energy at the bottom and back to potential energy on the opposite side, illustrating continuous energy exchange.
In a frictionless environment, the skateboarder would reach the same height on the opposite side because all potential energy would convert to kinetic and back without loss.
Friction transforms some kinetic energy into thermal energy, reducing the height the skateboarder reaches due to the dissipation of energy as heat.
Conceptual Question 1
Object sliding on a frictionless surface from point B.
Question: At what points does the object have the most KE (kinetic energy) and PE (potential energy?), exploring where energy is maximized based on position.
Example 1: Hurricane Tree Limb
A tree limb (mass = 22 kg, height = 13.3 m) falls on a roof 6 m above the ground.
A. What is the speed of the limb when it reaches the roof?
B. Find the kinetic energy of the limb when it reaches the roof.
Conceptual Question 2
Explain the energy transformation at different points in a diagram showing a roller coaster or similar setup, detailing the conversion between potential and kinetic energy.
Draw bar graphs showing Gravitational Potential Energy and Kinetic Energy at each point, illustrating how energy changes between potential and kinetic forms at each stage.
Example 2
A) What is the PE (Potential Energy), KE (Kinetic Energy), and Total Mechanical Energy of each point on a trajectory? Assume a = 10 m/s^2
B) What is h (height) at a specific point, requiring calculation based on energy principles.
Example 3: Roller Coaster
Frictionless roller coaster:
Initial speed (V_0) = 10.00 m/s
Initial height (h) = 100.0 m
Mass (m) = 1000.0 kg
(A) Speed at point A?
(B) Speed at point B?
(C) How high will it move up on the last hill?
Solution:
a) Height at A = height at Start. Both have some PE. ME = PE + KE. If KE is the same, speed is the same; V = 10 m/s
b) MEi = ME2
PE1 + KE1 = PE2 + KE2
PE = mgh
KE = \frac{1}{2}mv^2
(1000kg)(9.8 m/s^2)(100m) + \frac{1}{2}(1000 kg)(10m/s)^2 = (1000kg)(9.8 m/s^2)(\frac{100}{2}m) + \frac{1}{2}(1000kg)v^2
103000 = 49000 + \frac{1}{2}(1000kg)v^2
v = 32.86 m/s
c) PEi + KEi = PEc + KEc. At the highest point, V_f = 0 m/s
(1000kg)(9.8 m/s^2)(100m) + \frac{1}{2}(1000kg)(10m/s)^2 = (1000 kg)(9.8 m/s^2)h
103000 = 9800h
h = 105.10m
Example 4: Baseball
Baseball (mass = 0.86 kg) hit over the outfield fence.
Leaves bat at 36.0 m/s.
Caught 7.2 m above the point where it was hit.
Neglecting air resistance, what is the speed of the ball when caught?
Elastic Potential Energy
Energy stored by stretching or compressing an elastic object (like a spring) by an external force, representing potential to do work when released.
Hooke’s Law (Robert Hooke, 1678):
The force needed to extend or compress a spring is proportional to that distance, describing the linear relationship between force and displacement.
F_s = kx
k = spring constant (N/m) (stiffness of the spring), indicating how much force is needed to stretch or compress the spring by a unit length.
x = displacement (from equilibrium), measuring how far the spring is stretched or compressed from its relaxed position.
Elastic Potential Energy (EPE)
EPE = \frac{1}{2}kx^2
Conservation of Energy:
MEi = MEf
PEi + KEi + EPEi = PEf + KEf + EPEf
Example 5: Spring
A force of 120.0 N applied to a spring causes a stretch of 0.0225m.
A) What is the spring constant?
B) What is the potential energy of this spring when it is compressed by 0.0350m?
Example 6: Block and Spring
A 1.7 kg block slides on a horizontal, frictionless surface until it encounters a spring (k = 955 N/m).
The block compresses the spring a distance of 0.046 m.
A) Find the initial speed of the block.
B) If the spring was to compress greater than 0.046m, would this mean the block was going faster or slower? Explain.
Conceptual Question 4
Explain the energy transformation for each point in the diagrams involving a spring and a block, detailing how kinetic energy is converted into elastic potential energy and vice versa.
Compare PE, KE, and EPE at each point, detailing how energy shifts between these forms during the interaction between the block and spring.
Example 7: Spring and Block
A 2000 N/m spring is compressed 0.3 m against a wall with a 0.75 kg block ready to be released off the spring.
The block is released and it slides along a frictionless surface.
a. Calculate the elastic potential energy stored in the spring.
b. Calculate the speed of the block when it is released and loses contact with the spring.
Example 8: Ball and Spring
A ball of mass 2.00 kg is dropped from a height of 1.5 m onto a massless spring (equilibrium length 0.50 m).
The ball compresses the spring by 0.20 m by the time it comes to a stop.
a. Calculate the spring constant of the spring.
b. If the ball had more mass, how would this affect the compression of the spring when the ball comes to a stop?
Example 9: Hanging Spring
A spring hangs at rest from a support. If you suspend a 0.75 kg mass from the spring, it elongates 0.09 m.
a. Calculate the elastic potential energy.
b. Calculate the elongation if you put a 1.25 kg mass on the spring.
Conceptual Question 5
Explain the energy transformation for each of the points indicated in the diagrams starting from 1, describing the conversion of gravitational potential energy to kinetic energy and elastic potential energy.
Compare the PE, KE, and EPE at each of the points in the diagram, providing a comprehensive analysis of energy distribution and transformation processes.
Work
Definition: The ability to transfer energy, involving applying a force over a distance that results in displacement.
When a net force is exerted on an object while the object moves a certain distance, the object will be accelerated, and its velocity will increase.
Work has been done.
Units: Joules (J), same as energy!
Work - Directionality
If the force acts in the same direction as the object's motion, then the work done is positive, indicating energy is transferred to the object.
If the force acts in the opposite direction of the object's motion, then the work done is negative, indicating energy is transferred away from the object.
If the object does not move, then zero work is done, regardless of the force applied.
Formulas:
Work = Fdcos(\theta)
Work = Fd (when \theta = 0 degrees), indicating maximum energy transfer when force and displacement are aligned.
Example: Planet Orbiting the Sun
A force exerted perpendicular to the direction of motion, such as gravity acting on a planet in a circular orbit.
If the orbit is circular, then the force is always perpendicular to the direction of motion, meaning no work is done on the planet by gravity.
The speed of the planet doesn’t change, and so the right side of the equation is zero. Therefore, the work done is also zero, illustrating the conservation of kinetic energy.
Work-Energy Theorem
The net work done by the forces on an object equals the change in its kinetic energy, linking work and energy directly.
Net work is defined to be the sum of work done by all external forces (think of net force), representing the total energy transferred to or from the object.
W = \Delta KE
Where \Delta means “change in”
Conceptual Question 6
In which scenario is positive work being done? Zero work? Negative? Understanding the sign of work based on force and displacement.
A) Zero Work
B) Positive Work
C) Negative Work
Conceptual Question 7
Is there work for the following situations? Analyzing scenarios to determine if work is performed based on force and displacement.
A) A teacher applies a force to a wall and becomes exhausted. - Zero work because the displacement is 0, highlighting that effort without movement does not constitute work.
B) A book falls off a table and free falls to the ground. - The Force of Gravity does work on the book, making it go down, demonstrating gravitational force doing work.
C) A waiter carries a tray of food above his head with one arm and moves straight across the room at constant speed. - Zero work because the angle between the force and the displacement is 90 degrees, showing that perpendicular forces do no work.
D) A rocket accelerates straight up into the air. - The Force of the Thrust is providing positive work to the rocket. The force of gravity is doing negative work, indicating thrust adds kinetic energy while gravity opposes motion.
Example 11: Hockey Puck
A hockey player exerts a constant 4.5-N on a 0.105kg hockey puck which starts from rest and slides across the ice for a distance of 0.15m.
A) Draw a free body diagram of the example.
B) How much work does the player do on the puck?
C) How much work is done by the force of gravity?
D) How much work is done by the normal force?
E) What is the change in the puck’s kinetic energy? 0.675 J
Solution:
B) Wa = Fa d \cos \theta, \theta = 0^\circ (Force is in the same direction as the object's motion, so cos(0) = 1)
W_a = (4.5N)(0.15m)(\cos(0)) = 0.675 J
C) Wg = Fg d \cos \theta, \theta = 90^\circ (Force is perpendicular to the object's motion, so cos(90) = 0)
\cos(90^\circ) = 0, so W_g=0J
D) WN = FN d \cos \theta, \theta = 90^\circ (Force is perpendicular to the object's motion, so cos(90) = 0)
\cos(90^\circ) = 0, so W_N=0J
e) \Delta KE = KEf - KEi = \frac{1}{2}mvf^2 - \frac{1}{2}mvi^2
a = \frac{F}{m} = \frac{4.5}{0.105} = 42.86 m/s^2
v_i = 0 m/s
\Delta x = 0.15 m
vf^2 = vi^2 + 2a \Delta x
v_f^2 = 0 + 2(42.86 m/s^2)(0.15)
v_f = 3.59 m/s
\Delta KE = \frac{1}{2} m v_f^2 = \frac{1}{2} (0.105 kg)(3.59 m/s)^2 = 0.675 J
Example 12: Sailor Pulling a Boat
A sailor uses a rope to pull a 40 kg boat a distance of 30 m along a dock, making a 25 degree angle with the horizontal.
(A) How much work does the sailor do on the boat if he exerts a force of 255 N on the rope?
(B) What is the total work done on the boat?
(C) What is the speed of the boat after it has traveled 30 m if it starts with an initial velocity of 2 m/s?
Solution:
A) WA = Fi d \cos \alpha = (255N)(30m) \cos(25^\circ) = 6933.25J
B) W{tot} = WA + Wg + WN where Wg = 0 (perpendicular) and WN = 0 to motion of the boat.
W_{tot} = 6933.25J
C) W{tot} = \Delta KE = \frac{1}{2} m vf^2 + \frac{1}{2} m v_i^2
6933.25J = \frac{1}{2}(40kg) v_f^2 + \frac{1}{2} (40kg) (2 m/s)^2
v_f^2 = 18.73 m/s
Example 13: Boy Pulling a Sled
A boy exerts a force of 11N at 29 degrees above the horizontal on a 6.4-kg sled.
A) Find the work done by the boy if the sled moves 2 meters, assuming the sled starts with an initial speed of 0.5 m/s and slides horizontally without friction.
B) If the frictional force is 2N, what is the work done by the frictional force?
C) What is the total work done?
Solution:
A) WA = FA d \cos \theta = (11N)(2m)(\cos(29)) = 19.24 J
B) WF = FF d \cos \theta = (2N)(2m) \cos(180) = -4 J
C) W{tot} = WA + WF + Wg + W_N = 19.24 - 4 = 15.24 J
Example 14: Bicycle Rider Pushing Bike Up a Hill
A bicycle rider pushes a bicycle that has a mass of 1.3-kg up a steep hill. The incline is 25 degrees and the road is 275m long. The rider pushes the bike parallel to the road with a force of 25-N. (Neglect Friction)
(A) How much work is done by the normal force on the bike?
(B) How much work is done by the force of gravity?
(C) How much work is done by the rider?
(D) What is the Total Work?
(E) What is the final speed of the bike at the top of the hill?
(F) If the incline plane was steeper would this require the rider to do more or less work? Explain.
Solution:
A) WN = FN d \cos \theta. FN is perpendicular to the motion. Where \theta = 90^\circ. \cos(90^\circ) = 0 so WN = 0
B) Wg = Fg \cos \theta. Where F_g = mg = (1.3)(9.8) = 12.74 N. \theta = 90^\circ + 25 = 115^\circ.
W_g = (12.74N)(275m)(\cos(115)) = -1480.64 J
C) WF = FA d \cos \theta = (25N)(275m)(\cos(0)) = 6875 J
D) W{tot} = WA + Wg + WN = 6875 - 1480 = 5364.14 J
E) W{tot} = \Delta KE = \frac{1}{2} m vi^2. 5364.14 = \frac{1}{2} (1.3) v_f^2
v_f = 90.84 m/s
Work stretching a spring Example
(A) It takes 40 J of work to stretch a spring from 0.05 m elongation to a 0.06 m elongation. How much work is needed to stretch it from a 0.06 m elongation to a 0.07 m elongation?
Power
Definition
The rate at which work is done. Work causes a change in energy. The rate at which the energy is transformed is power. In other words, power is the change in energy over time.
Power = \frac{Work}{time}
P = \frac{W}{t}
Units: Watts (W) or J/s
Also
P = Fv
Example 18: Electric Motor Lifting an Elevator
An electric motor lifts an elevator 9m in 15s by exerting an upward force of 12000N. What power does the motor produce?
Example 19: Car Passing a Truck
To pass a slow-moving truck, you need your 1300kg car to accelerate from 13.4 m/s to 17.9 m/s in 3 seconds. What is the power required to pass?
Example 20: Car Moving Up a Slope
It takes a force of 1280N to keep a 1500kg car moving with constant speed up a slope of 5 degrees. If the engine delivers 37,300 Watts, what is the speed of the car?