# Topics in textbook Actual Work Definition of Work Potential Energy

## Presentation on theme: "Topics in textbook Actual Work Definition of Work Potential Energy"— Presentation transcript:

Topics in textbook Actual Work Definition of Work Potential Energy
Work on Particle Potential Energy Definition of Energy Work-Energy Equation Kinetics Energy Definition of Energs

Three approaches for solving dynamics
Newton’s 2nd Law Kinematics Eq: path 1) Direct Method A From 2nd Law (kinetics Eq) work (and potential energy) of Force i along the path 2) Work and Energy From 2nd Law 3) Impulse and Momentum linear impulse of Force i From 2nd Law

Work and Energy Usually convenient From 2nd Law change of
Newton’s 2nd Law Work and Energy Usually convenient when F = F(s), and you want to find velocity at final state (without finding acc. first). From 2nd Law change of kinetic energy kinetic energy at B path B Work of Force i along the path kinetic energy at A A Work of a force during small displacement Principle of work and Energy kinetic energy

Work & Energy Work Energy “something” successful done by force.
The objective of force is to create the movement. a Work [Unit: Newton- Meter (J: joule) ] “something” successful done by force. - it depends on and ( “indicator” of successiveness of force ) - “Indicator” has sign Energy - Capacity/Potential to do the work. ( the work is not yet done)

Work done on Particle Work by a force Work done over particle A
path P (inactive force) Work done over particle A Since , the total work done on object is …… sum of works done by all forces over the particle A

Note on work is positive when has the same direction.
path is positive when has the same direction. Unit of work is N-m or Joule (J). Active force is the force that does the work Reactive force = constrain force that does not do the work

A (spring stretched length is 0.5m)
The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. The block is not tipping. B A (spring stretched length is 0.5m) Pos B Horizontal Force P: constant Pos A Weight W: constant. “Active Force” Spring Force Fs.: varying Force Normal Force NB : constant “Inactive Force”

The 10-kg block rest on a smooth incline
The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. P = 400 N constant along the path scalar calculation (aware of +/- ) Spring = k(0.5+x) x=0 to 0.5 mg = (10)(9.81) constant along the path mg Does not do the work P =400N Spring Normal Force N This force does no work since it is always perpendicular to the displacement.

Work done on Particle t Work done over particle A path sum of works
done by all forces over the particle A P Pos B Pos A Work done on particle P during path A->B, is to increase kinetic energy of particle

Work and Energy Usually convenient From 2nd Law change of
Newton’s 2nd Law Work and Energy Usually convenient when F = F(s), and you want to find velocity at final state (without finding acc. first). From 2nd Law change of kinetic energy kinetic energy at B path B Work of Force i along the path kinetic energy at A A Work of a force during small displacement Principle of work and Energy kinetic energy

path A T is the work done on a particle to accelerate it from rest to the velocity v Principle of work and Energy Unit of T is N-m or Joule (J) Advantage Scalar equation. (1 unknown) Integral Equation (not instantaneous eq like 2nd Law) No need to find acceleration first Get change in velocity directly from active forces. it can be applied to system of particles with frictionless and non-deformable links

Motivation : Work-Energy Equation
Solve for every instant t (Newtonian Equation is instantaneous Equation) We have to write FBD of general position of time t initial final New way: Use Work-Energy Equation

SP3/3 The block A is released from rest, and pulls the 200-kg log up
SP3/3 The block A is released from rest, and pulls the 200-kg log up. Determine the velocity of the block A when it hits the ground Static? The pulley C is moving T T 2T 2T 200(g) T N 125(g) constant at any time t A

Three approaches for solving dynamics
Newton’s 2nd Law Kinematics: Direct Method From 2nd Law work (and potential energy) of Force i along the path Work and Energy From 2nd Law Impulse and Momentum linear impulse of Force i From 2nd Law

Principle of work and Energy Advantage of Work-Energy Eq. No need to find acceleration first Work-Energy Eq. Get change in velocity directly from active forces. can be applied to system of particle using frictionless and non-deformable links path A Usually convenient when F = F(s)

How to calculate Work In general In xyz-coord In nt-coord In rq-coord
path In general displacement x force component in the direction of displacement In xyz-coord scalar (be careful Of +/-) displacement in the direction of force x force component In nt-coord In rq-coord

M3/107) Calculate the work done on 10-kg object with the constant Force ( F= 8N ) during the curve path AB. x-y y x mg F=8N (const) N Does not do the work Ans If F is not constant, how to calculate it?

If F is not constant y x More general case or

M3/107) Calculate the work done by F during the curve path AB.
n-t Fn doest not effect works! s engine thrust a

M3/107) Calculate the work done by F during the curve path AB.
r-q r-q reference point central force

M3/107) Calculate the work done by F during the curve path AB.
n-t co-ord s N engine thrust central force r-q reference point

M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 5-N tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B. mg F Work-Energy Eq. general position N Does not do the work r-q coordinate reference point

Hibbeler Ex. 14.1 Given: x1 = 0.5 m P = 400 N s = 2 m
Find: Total work done by all forces Plan: 1) Draw a FBD 2) Calculate for work done by each forces (Horizontal P, Spring force Fs, Weight W, and Normal force N). Then sum all works together.

Hibbeler Ex (cont.)

Hibbeler Ex (cont.)

Hibbeler Ex (cont.)

Work and Curvilinear Motion

No need to find acceleration first Get change in velocity directly from active forces. Work-Energy Eq. is a scalar equation. it can be applied to system of particles using frictionless and non-deformable links

Work on frictionless connected particles
Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.) internal force R and –R will have the same displacement. So, the sum of these works are zero.

initial state The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant) Final state N causes no work! System selection is not so good (you have to calculate Tension T for its work)

M3/131) The ball is released from position A with a velocity of 3m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calcuate the velocity v of the ball as it passes position C. Work-Energy Eq. T does no work system mg r-q coordinate reference point system

Power Power is defined as time rate of work
path Power is defined as time rate of work A (scalar quantity) For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) Unit of power: Watt (W) = J/s = N-m/s

Mechanical Efficiency
If energy applied to the machine occurs during the same time interval at which it is removed. Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

A car has a mass of 2 Mg and an engine efficiency of e = 0. 65
A car has a mass of 2 Mg and an engine efficiency of e = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a x Constant acceleration:

A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. v=? F = 30 N (const) Energy Approach 2F= 2(30) 50 N B 50N s = 10 m (start from rest)

summation of all forces
Work and Energy kinetic energy at B path B kinetic energy at A A summation of all forces We found that …. It is much easier to solve dynamic problem, if we think the work done by spring and gravity force in the form of Potential Energy Work from Gravity Force Work from spring Work from all other forces (not spring & gravitation) Gravitational Potential Energy Elastic Potential Energy kinetic energy

Work of Gravity Force Only depends on position at final state (2) Only depends on position at initial state (2) > 0 any path 1 energy level (higher) W=mg Work done by W , only depends on the initial state position and final state position only, i.e. , it does not depends on actual path 2 energy level (lower) h Think in Term of “Potential Energy” (for convenience) Fixed reference line point function Potential Energy - Energy from gravity field Work done by Gravity Force: from position 1 to position 2 Energy “Emission”: from position 1 to position 2 Work = “Energy in Transfer” Work by couple is - positive if M has the same sense as dq - negative if M has the inverse sense as dq Work done over Object 43

when change in g is significant
Define as negative of work done from the position to the potential energy at r is from

position at initial state (1)
(direction may be opposite) Work of Spring Force Only depends on position at final state (2) Only depends on position at initial state (1) 1 L natural length (unstretched length) Work done by Spring , depends only on the initial state and final state only, i.e. , it does not depends on actual path 2 any path x : distance , stretched or compressed from natural length Think in Term of “Energy” (for convenience) point function Work done by Spring Force: from position 1 to position 2 Energy Emission: from position 1 to position 2 Work done by Gravity Force: from position 1 to position 2 Energy Emission: from position 1 to position 2

Work-Energy Equation FBD ** FBD N N Work-Energy Equation
(Use Energy Concept) FBD Not Recommended Method in this course N N (Conservative Force) Think of Energy Work-Energy Equation (2nd Form) Work-Energy Equation (1st Form) Virtual work by non-conservative forces. Energy Concept final location initial location

Work-Energy Equation (1st Form) FBD Work-Energy Equation (2nd Form) N (Conservative Force) Think of Energy The virtual work done by external active forces (other than gravitational and spring forces accounted for in the potential energy terms) on an ideal mechanical system in equilibrium, equals the corresponding change in the total elastic and gravitational potential energy of the system for any and all virtual displacements consistent with the constraints.

M3/173) The 0.6-kg slider is released from rest at A and slides down under the influence of its own weight and of the spring of k = 120 N/m. Determine the speed of the slider and the normal force at point B. The unstrecthed length of the spring is 200 mm. gravitational potential datum At position B gravitational Potential datum

Hibbeler Ex (cont.)

H14/92) The 1-kg collar has a speed of 1. 5 m/s at A
H14/92) The 1-kg collar has a speed of 1.5 m/s at A. the attached spring has an unstretched length of 0.6 m and a stiffness of k = 150 N/m. When it reaches point B, determine its velocity, the rate of decrease in its speed, and the normal force of the collar to the rod. mg N F mg F=kx N

M3/172) The cars of an amusement-park ride have a speed v=90 km/h at the lowest part of the track. Determine their speed v at the highest part of the track. Neglect energy loss due to friction. CFO system gravitational potential datum

Hibbeler Ex (cont.) Datum an = v2/,  = 20 m

Integral Equation (not instantaneous equation like 2nd Law)
Advantage Integral Equation (not instantaneous equation like 2nd Law) Scalar equation. (easy to handle with1 unknown) Get change in velocity directly. (No need to find acceleration first) Handle with only active forces. it can be applied to system of particles with frictionless and non-deformable links We will see this later, when applying at system of particles

Work on frictionless connected particles
B C O Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.) internal force R and –R will have the same displacement. So, the sum of these works are zero.

Work on frictionless connected particles
Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.) internal force R and –R will have the same displacement. So, the sum of these works are zero.

initial state Final state The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant) T T T T F F Object A Object B system T is internal force F (excluding from Work Calculation) We have no interest in T, thus object separation (separating object A and B) is not good in this problem.

(Active Force Diagram )
Internal forces (action/reaction pair) are cancelled with each other. Work on Connected Rigid Bodies AFD (Active Force Diagram ) Active Force: Forces that do work. A B C O Constrain Force (reactive force) causes no work because the movement is not possible. friction-less system AFD

Work on Connected Rigid Bodies
Internal forces (action/reaction pair) are cancelled with each other. Work on Connected Rigid Bodies AFD (Active Force Diagram ) Active Force: Forces that do work. A B C O Constrain Force (reactive force) causes no work because the movement is not possible. friction-less system AFD

(Active Force Diagram )
Internal forces (action/reaction pair) are cancelled with each other. Work on Connected Rigid Bodies AFD (Active Force Diagram ) Active Force: Forces that do work. A B C O Constrain Force (reactive force) causes no work because the movement is not possible. friction-less system AFD

M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. System: block A + block B + cord+ 2 Pulleys up Position A: at rest (assume) Position B: block B moves down as 1 meter datum unsolvable

M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. System: block A + block B + cord+ 2 Pulleys up Position A: at rest (assume) Position B: block B moves down as 1 meter datum

H14/16) Block A rest on a surface which has friction
H14/16) Block A rest on a surface which has friction. Determine the distance d cylinder B must move down so that A has a speed of starting from rest. 20 N System: block A + block B + cord+ 2 Pulleys 50 N Position A: at rest Position B: block B moves down as d meter

3/168) The system is released from rest with q=180, where the uncompressed spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar. Determine the angle q corresponding to the maximum spring compression. O2-1 O2-2 O1 r L r datum System: O1+O2+O3+4 rods Position A: at rest with q=180 Position B: maximum compression

Power Power is defined as time rate of work
path Power is defined as time rate of work A (scalar quantity) For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) Unit of power: Watt (W) = J/s = N-m/s

Mechanical Efficiency
If energy applied to the machine occurs during the same time interval at which it is removed. Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. v=? No! F = 30 N (const) 50N 2F= 2(30) Energy Approach s = 10 m (start from rest) F = 30 N (const) 50 N 50 N

A car has a mass of 2 Mg and an engine efficiency of e = 0. 65
A car has a mass of 2 Mg and an engine efficiency of e = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a x Constant acceleration:

Summary Make sure you write FBD (no FBD, no score)
Scalar Equation (Only 1 unknown) Equation itself is not hard to solve, but calculating work may be more difficult than you thought.

Recommended Problem M3/144 M3/155 M3/160 M3/168 H14/93 , H 14/96

Download ppt "Topics in textbook Actual Work Definition of Work Potential Energy"

Similar presentations