Kinetic Energy and Work

Kinetic Energy and Work
Chapter 7 Kinetic Energy and Work

Introduction to Energy
The concept of energy is one of the most important topics in science Every physical process that occurs in the Universe involves energy and energy transfers or transformations

Work and energy are scalars, measured in N·m or Joules, J, 1J=kg∙m2/s2
Energy: scalar quantity associated with a state (or condition) of one or more objects. Work and energy are scalars, measured in N·m or Joules, J, 1J=kg∙m2/s2 Energy can exist in many forms - mechanical, electrical, nuclear, thermal, chemical….

Energy Approach to Problems
The energy approach to describing motion is particularly useful when the force is not constant An approach will involve Conservation of Energy This could be extended to biological organisms, technological systems and engineering situations

Energy can be converted from one type to another but never destroyed.
Work and Energy Energy is a conserved quantity - the total amount of energy in the universe is constant. Energy can be converted from one type to another but never destroyed. Work and energy concepts can simplify solutions of mechanical problems - they can be used in an alternative analysis

Systems A system is a small portion of the Universe
We will ignore the details of the rest of the Universe A critical skill is to identify the system

Valid System A valid system may be a single object or particle
be a collection of objects or particles be a region of space vary in size and shape

Environment There is a system boundary around the system
The boundary is an imaginary surface It does not necessarily correspond to a physical boundary The boundary divides the system from the environment The environment is the rest of the Universe

Work The work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude, F, of the force, the magnitude Δr of the displacement of the point of application of the force, and cos θ, where θ is the angle between the force and the displacement vectors

Work W = F Δr cos θ The displacement is that of the point of application of the force A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application

Work Example cos θ = cos 90° = 0
The normal force, n, and the gravitational force, mg, do no work on the object cos θ = cos 90° = 0 The force F does do work on the object

More About Work The system and the environment must be determined when dealing with work The environment does work on the system (Work by the environment on the system) The sign of the work depends on the direction of F relative to Δr Work is positive when projection of F onto Δr is in the same direction as the displacement Work is negative when the projection is in the opposite direction

Work Is An Energy Transfer
This is important for a system approach to solving a problem If the work is done on a system and it is positive, energy is transferred to the system If the work done on the system is negative, energy is transferred from the system

Work done by a constant force
Work done on an object by a constant force is defined to be the product of the magnitude of the displacement and the component of the force parallel to the displacement Where FII is the component of the force F parallel to the displacement d W = FII · d

Work 1J=kg∙m2/s2 In other words - Where θ is the angle between F and d
If θ is > 90o, work is negative. A decelerating car has negative work done on it by its engine. The unit of work is called Joule (J), 1 J = 1 N·m 1J=kg∙m2/s2 W = F d cosq q d F

Scalar Product of Two Vectors
The scalar product of two vectors is written as A . B It is also called the dot product A . B = A B cos θ θ is the angle between A and B

Scalar Product The scalar product is commutative A . B = B . A
The scalar product obeys the distributive law of multiplication A . (B + C) = A . B + A . C

Dot Products of Unit Vectors
Using component form with A and B:

Work - on and by A person pushes block 30 m along the ground by exerting force of 25 N on the trolley. How much work does the person do on the trolley? W = F d = 25N x 30m = 750 Nm Trolley does -750 Nm work on the person Ftp Fpt d

A force acts on an object as the object moves in the x direction from the origin to x = 5.00 m. Find the work W = Fdr done on the object by the force.

Mechanical Energy Mechanical energy (energy associated with masses) can be thought of as having two components: kinetic and potential Kinetic energy is energy of motion Potential energy is energy of position

Kinetic Energy In many situations, energy can be considered as “the ability to do work” Energy can exist in different forms Kinetic energy is the energy associated with the motion of an object A moving object can do work on another object E.g hammer does work on nail.

Kinetic Energy Consider an object with mass m moving in a straight line with initial velocity vi. To accelerate it uniformly to a speed vf a constant net force F is exerted on it parallel to motion over a distance d. Work done on object W = F d = m a d (NII) So

Kinetic Energy If we rearrange this we obtain We define the quantity ½mv2 to be the translational kinetic energy (KE) of the object This is the ‘Work-Energy Theorem’: “The net work done on an object is equal to its change in kinetic energy” W = DKE

The Work-Energy Theorem
W = DKE Note The net work done on an object is the work done by the net force. Units of energy and work must be the same (J)

Energy: scalar quantity associated with a state (or condition) of one or more objects.
Kinetic energy. Work. Work - Kinetic energy theorem. Work done by a constant force - Gravitational force V. Work done by a variable force. - Spring force. 1D-Analysis 3D-Analysis VI. Power

I. Kinetic energy II. Work Units:
Energy associated with the state of motion of an object. Units: 1 Joule = 1J = 1 kgm2/s2 II. Work To  +W From  -W Energy transferred “to” or “from” an object by means of a force acting on the object.

- Constant force: Work done by the force = Energy transfer due to the force.

To calculate the work done on an object by a force during
a displacement, we use only the force component along the object’s displacement. The force component perpendicular to the displacement does zero work. Assumptions: 1) F = constant force 2) Object is particle-like (rigid object, all parts of the object must move together). A force does +W when it has a vector component in the same direction as the displacement, and –W when it has a vector component in the opposite direction. W=0 when it has no such vector component.

Net work done by several forces = Sum of works done by
individual forces. Calculation: 1) Wnet= W1+W2+W3+… 2) Fnet  Wnet=Fnet d

II. Work-Kinetic Energy Theorem III. Work done by a constant force
Change in the kinetic energy of the particle = Net work done on the particle III. Work done by a constant force - Gravitational force: Rising object: W= mgd cos180º = -mgd  Fg transfers mgd energy from the object’s kinetic energy. Falling down object: W= mgd cos 0º = +mgd  Fg transfers mgd energy to the object’s kinetic energy.

External applied force + Gravitational force:
Object stationary before and after the lift: Wa+Wg=0 The applied force transfers the same amount of energy to the object as the gravitational force transfers from the object.

IV. Work done by a variable force
- Spring force: Hooke’s law k = spring constant  measures spring’s stiffness. Units: N/m

Hooke’s Law When x is positive (spring is stretched), F is negative
When x is 0 (at the equilibrium position), F is 0 When x is negative (spring is compressed), F is positive

Hooke’s Law The force exerted by the spring is always directed opposite to the displacement from equilibrium F is called the restoring force If the block is released it will oscillate back and forth between –x and x

Work done by a spring force:
Hooke’s law: xi Δx x Work done by a spring force: Spring is massless  mspring << mblock Ideal spring  obeys Hooke’s law exactly. Contact between the block and floor is frictionless. Block is particle-like. Assumptions:

The block displacement must be divided
- Calculation: The block displacement must be divided into many segments of infinitesimal width, Δx. 2) F(x) ≈ cte within each short Δx segment. Fx xi Δx xf x Fj Ws>0  Block ends up closer to the relaxed position (x=0) than it was initially. Ws<0  Block ends up further away from x=0. Ws=0  Block ends up at x=0.

Work done by an applied force + spring force:
Block stationary before and after the displacement: Wa= -Ws  The work done by the applied force displacing the block is the negative of the work done by the spring force.

If it takes 4. 00 J of work to stretch a Hooke's-law spring 10
If it takes 4.00 J of work to stretch a Hooke's-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm.

A 2. 00-kg block is attached to a spring of force constant 500 N/m
A 2.00-kg block is attached to a spring of force constant 500 N/m. The block is pulled 5.00 cm to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is

Work done by a general variable force:
Assume that during a very small displacement, Δx, F is constant For that displacement, W ~ F Δx For all of the intervals,

Work done by a general variable force:
Therefore, The work done is equal to the area under the curve

Work-Kinetic Energy Theorem - Variable force
3D-Analysis Work-Kinetic Energy Theorem - Variable force

V. Power Time rate at which the applied force does work.
- Average power: amount of work done in an amount of time Δt by a force. Instantaneous power: instantaneous time rate of doing work. Units: 1 watt= 1 W = 1J/s 1 kilowatt-hour = 1000W·h = 1000J/s x 3600s = 3.6 x 106 J = 3.6 MJ F φ x

In the figure below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across a frictionless floor. Find an expression for the speed vf at the end of that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (a) N mg Fx Fy N Fx Fy mg

(c) The situation in fig
(c) The situation in fig.(b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance. N mg Fx Fy N Fx Fy mg

A small particle of mass m is pulled to the top of a frictionless half-cylinder (of radius R) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that F = mgcos. (Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating W = Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder.

A small particle of mass m is pulled to the top of a frictionless half-cylinder (of radius R) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that F = mgcos. (Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating W = Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder. (a)

A small particle of mass m is pulled to the top of a frictionless half-cylinder (of radius R) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that F = mgcos. (Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating W = Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder. (b) (We use radian measure to express the next bit of displacement dr as in terms of the next bit of angle moved through dθ: dr=Rdθ)

Two springs with negligible masses, one with spring constant k1 and the other with spring constant k2, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. When the glider is in equilibrium, spring 1 is stretched by extension xi1 to the right of its unstretched length and spring 2 is stretched by xi2 to the left. Now a horizontal force Fapp is applied to the glider to move it a distance xa to the right from its equilibrium position. Show that in this process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b) the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by the force Fapp is (k1 + k2)xa2.

Two springs with negligible masses, one with spring constant k1 and the other with spring constant k2, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. When the glider is in equilibrium, spring 1 is stretched by extension xi1 to the right of its unstretched length and spring 2 is stretched by xi2 to the left. Now a horizontal force Fapp is applied to the glider to move it a distance xa to the right from its equilibrium position. Show that in this process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b) the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by the force Fapp is (k1 + k2)xa2. (a) (b)

Two springs with negligible masses, one with spring constant k1 and the other with spring constant k2, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. When the glider is in equilibrium, spring 1 is stretched by extension xi1 to the right of its unstretched length and spring 2 is stretched by xi2 to the left. Now a horizontal force Fapp is applied to the glider to move it a distance xa to the right from its equilibrium position. Show that in this process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b) the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by the force Fapp is (k1 + k2)xa2. (c) Before the horizontal force is applied, the springs exert equal forces:

Two springs with negligible masses, one with spring constant k1 and the other with spring constant k2, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. When the glider is in equilibrium, spring 1 is stretched by extension xi1 to the right of its unstretched length and spring 2 is stretched by xi2 to the left. Now a horizontal force Fapp is applied to the glider to move it a distance xa to the right from its equilibrium position. Show that in this process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b) the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by the force Fapp is (k1 + k2)xa2. (d)

N6. A 2kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s?

N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x, Fig. below. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s?

N12. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kg book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net force done on the book by Fa, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement? x y mg N Fgy Fgx

N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net force done on the book by Fa, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement? y x mg N Fgy Fgx

N15. (a) Estimate the work done represented by the graph below
in displacing the particle from x=1 to x=3m. (b) The curve is given by F=a/x2, with a=9Nm2. Calculate the work using integration

N19. An elevator has a mass of 4500kg and can carry a maximum
load of 1800kg. If the cab is moving upward at full load at constant speed 3.8m/s, what power is required of the force moving the cab to maintain that speed? Fa mg

N17. A single force acts on a body that moves along an x-axis.
The figure below shows the velocity component versus time for the body. For each of the intervals AB, BC, CD, and DE, give the sign (plus or minus) of the work done by the force, or state that the work is zero. v B C D A t E

15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister? P2 T T T P1 mg

15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister? P2 T T T P1 mg (b) To rise “m” 0.02m, two segments of the cord must be shorten by that amount. Thus, the amount of the string pulled down at the left end is: 0.04m

There is no change in kinetic energy.
15E. In the figure below, a cord runs around two massless, frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister? P2 T T T P1 mg WF+WFg=0 There is no change in kinetic energy.

From the point of view of kinetics, does the relative position of the
Challenging problems – Chapter 7 Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. From the point of view of kinetics, does the relative position of the trolleys matter? If the rod can only stand an applied force of 500N, which trolley should be up front? N1 Situation 1 N2 F1r F2r F m1g m2g

From the point of view of kinetics, does the relative position of the
Challenging problems – Chapter 7 Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. From the point of view of kinetics, does the relative position of the trolleys matter? If the rod can only stand an applied force of 500N, which trolley should be up front? N1 Situation 1 N2 F1r F2r F m1g m2g Action and reaction forces: F1r force that the rod does on m1 F2r force that the rod does on m2 F’2r F’1r F’1r force that m1 does on rod. F’2r force that m2 does on rod. Situation 1 Rod negiglible mass, Fnet on rod=0 F’1r = F’2r  rigid rod does not deform F1r = F’1r  Action / Reaction Forces that the rod does on the blocks as it tries to counteract the deformation that F could induce.

Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. From the point of view of kinetics, does the relative position of the trolleys matter? If the rod can only stand an applied force of 500N, which trolley should be up front? F’2r F’1r Action and reaction forces: F1r force that the rod does on m1 F2r force that the rod does on m2 Situation 2 F’1r force that m1 does on rod. F’2r force that m2 does on rod. Rod negiglible mass, Fnet on rod=0 F’1r = F’2r  rigid rod does not deform F1r = F’1r  Action / Reaction Situation 2 N2 N1 From kinetic point of view, Situation1=Situation2same acceleration”. F2r F1r F From dynamics  F1r(1)≠F1r(2) Only situation (1) is possible F1r=F’1r = 400N <Fmax=500N m1g m2g

2. A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces? (a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the car is traveling on a 100 inclined hill at 36km/h?

2. A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces? (a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the car is traveling on a 100 inclined hill at 36km/h? N f F Situation 1 mg

2. A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: (a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the car is traveling on a 100 inclined hill at 36km/h? Situation 2 N F Fgx f Fgy mg

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