 Energy, Work and Simple Machines  Chapter 10  Physics

Objectives: The student will be able to: 1. distinguish between work in the scientific sense as compared to the colloquial sense. 2. write the definition of work in terms of force and displacement and calculate the work done by a constant force when the force and displacement vectors are at an angle. 3. state and apply the relationship that work done with no opposing force equals the change in kinetic energy. 4.state and apply the relationship that work done against gravity equals the change in gravitational potential energy. 5.Define and calculate power from calculating the amount of work done by an object.

 What is work?  In science, the word work has a different meaning than you may be familiar with.  The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)

 Work  Work has its own meaning in physics.  Work is done on an object when an applied force acting on the object moves the object over a distance.  Work depends on two factors.  Force (F)  Displacement (d)

 Work  Work = Force x Displacement  W = Fd  Unit for Work = Newton Meter (Nm)  1 Nm = 1 Joule (J) (Same as Energy)  Work is a scalar quantity (no direction)  In doing Work the Displacement has to be in the same direction as the Force!

 Work or Not?  According to the scientific definition, what is work and what is not?  A teacher lecturing to her class  A mouse pushing a piece of cheese with its nose across the floor

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8 What’s work?  A scientist delivers a speech to an audience of his peers.  No  A body builder lifts 350 pounds above his head.  Yes  A mother carries her baby from room to room.  No  A father pushes a baby in a carriage.  Yes  A woman carries a 20 km grocery bag to her car?  No

Physicist ’ s definition of “ work ” dist Work = F x dist ∥ A scalar (not a vector) dist ∥

Scalar Dot Product? A product is obviously a result of multiplying 2 numbers. A scalar is a quantity with NO DIRECTION. So basically Work is found by multiplying the Force times the displacement and result is ENERGY, which has no direction associated with it. A dot product is basically a CONSTRAINT on the formula. In this case it means that F and x MUST be parallel. To ensure that they are parallel we add the cosine on the end. FORCE Displacement

Work FORCE Displacement

Work Done by a Constant Force The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement:

Work In the figure above, we see the woman applying a force at an angle theta. Only the HORIZONTAL COMPONENT actually causes the box to move and thus imparts energy to the box. The vertical component (Fsin  ) does NO work on the box because it is NOT parallel to the displacement.

Work Relates force to change in energy Scalar quantity Independent of time

Atlas holds up the Earth But he doesn ’ t move, dist ∥ = 0 Work= F x dist ∥ = 0 He doesn’t do any work!

 Work  Work can be Zero (W NET = 0) in three ways;  d = 0 (does not move or finishes where it starts)  F NET = 0 (v = 0 or v = constant)  F NET is perpendicular to d (F | d)

 Work  If F ll d, then W = Fd or W = -Fd  W = Fd, Force in same direction as displacement (  = 0 o : cos  = 1, Positive Work)  W = -Fd, Force is in the opposite direction as the displacement (  = 180 o : cos  = -1, Negative Work)

 Work From a Force vs Displacement Graph  If you have a Force vs Displacement Graph, where the Force is in Newtons (N) and the Displacement is in Meters (m), you can find the Work by finding the Area Under the Curve!  W = Area under F vs d graph

 Formula for Work  Work = Force x Distance The unit of force is newtons The unit of distance is meters The unit of work is newton-meters One newton-meter is equal to one joule So, the unit of work is a joule

Work Done by a Constant Force In the SI system, the units of work are joules: As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. SI unit = Joule 1 J = 1 N  m = 1 kg  m 2 /s 2

Work FORCE Displacement

Work can be positive or negative Man does positive work lifting box Man does negative work lowering box Gravity does positive work when box lowers Gravity does negative work when box is raised

Work performed climbing stairs lWork = Fd lForce l Subject weight lFrom mass, ie 65 kg lDisplacement l Height of each step lTypical 8 inches (20cm) lWork per step l 650N x 0.2 m = Nm lMultiply by the number of steps

Power lThe rate of doing work l Work = Fd Units: Fd/s = J/s = watt

 Energy  Energy is the property that describes an object’s ability to change itself or the environment around it.  Energy can be found in many forms.  Kinetic Energy (KE) – energy of motion.  Potential Energy (PE) – energy gained by a change in position or structure

 Kinetic Energy (KE)  Moving objects possess Kinetic Energy.  KE = ½ mv 2  Energy is a scalar quantity and has the unit of Joule (J) (1 J = 1 Nm)

Work done against gravity W = mgh Height object raised (m) Gravity (m/sec 2 ) Work (joules) Mass (g) Hewitt – Work and PE

The formula for kinetic energy A force (F) is applied to mass (m) and creates acceleration (a). After a distance (d), the ball has reached speed (v), therefore the work done is its mass times acceleration time distance: –W= fd = (ma) x d = mad –Also: d = ½ at 2 Replace d in the equation for work, combine similar terms: –W= ma (½ at 2 ) = ½ ma 2 t 2 –Also: v = at, so v 2 = a 2 t 2 Replace a 2 t 2 by v 2 shows that the resulting work is the formula for kinetic energy: –W = ½ mv 2 Hewitt – PE and KE Period 3 starts here

Work-Kinetic Energy Theorem Work-kinetic Energy Theorem: Net work done on a particle equals the change in its kinetic energy (KE) W = ΔKE

 Practice Questions  Explain who is doing more work and why:  A bricklayer carrying bricks and placing them on the wall of a building being constructed  Or a project supervisor observing and recording the progress of the workers from an observation booth.

 Practice Question  Explain who is doing more work and why:  A bricklayer carrying bricks and placing them on the wall of a building being constructed  Or a project supervisor observing and recording the progress of the workers from an observation booth.  Work is defined as a force applied to an object, moving that object a distance in the direction of the applied force. The bricklayer is doing more work.

 Practice Question  How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then pushing it back to its original position? How much power is used if this work is done in 20 seconds?

 Practice Question  How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then pushing it back to its original position? How much power is used if this work is done in 20 seconds?  Work = 7.0m x 50N x 2 = 700 Nm or 700 J  Power = 700J / 20s = 35 W

 Practice Question  How much power will it take to move a 10 kg mass at an acceleration of 2 m/s 2 a distance of 10 meters in 5 seconds? This problem requires you to use the formulas for force, work, and power all in the correct order.

 Practice Question  How much power will it take to move a 10 kg mass at an acceleration of 2 m/s 2 a distance of 10 meters in 5 seconds? This problem requires you to use the formulas for force, work, and power all in the correct order.  F = ma  F = 10 kg x 2 m/s = 20 N  W = Fd  W = 20 N x 10 m = 200 J  P = W/t  P = 200 J / 5 s = 40 watts

Calculate work done against gravity A crane lifts a steel beam with a mass of 1,500 kg. Calculate how much work is done against gravity if the beam is lifted 50 meters in the air. How much time does it take to lift the beam if the motor of the crane can do 10,000 joules of work per second? Practice Question

 You are asked for the work and time it takes to do work.  You are given mass, height, and work done per second.  Use: W = mgh.  Solve: W = (1,500 kg) ( 9.8 N/kg) (50 m) = 735,000 J  At a rate of 10,000 J/s, it takes 73.5 s to lift the beam. Calculate work done against gravity A crane lifts a steel beam with a mass of 1,500 kg. Calculate how much work is done against gravity if the beam is lifted 50 meters in the air. How much time does it take to lift the beam if the motor of the crane can do 10,000 joules of work per second? Practice Question

Calculating kinetic energy A car with a mass of 1,300 kg is going straight ahead at a speed of 30 m/s (67 mph). The brakes can supply a force of 9,500 N. Calculate: a) The kinetic energy of the car. b) The distance it takes to stop. Practice Question

 You are asked for kinetic energy and stopping distance  You are given mass, speed and force of brakes.  Use E k = 1 / 2 mv 2 and W= fd  Solve for E k = ½ (1,300 kg) ( 30 m/s) 2 = 585,000 J –To stop the car, work done by brakes = E k of car, so W = E k –Solve for distance = W ÷ f = 585,000J ÷ 9,500 N = 62 m Calculating kinetic energy A car with a mass of 1,300 kg is going straight ahead at a speed of 30 m/s (67 mph). The brakes can supply a force of 9,500 N. Calculate: a) The kinetic energy of the car. b) The distance it takes to stop. Practice Question

Elaboration Force, Distance, and Work – Trans 10 Work and Energy Internet Activity P. 261 #s1, 2, 3 P. 262 #s 4, 5, 6, 7, 8 P. 264 # 10 and 12 Questions and Problems Work and Power Stair Climbing and Power lab Section Review #s 15-21

Closure Kahoot 10.1 Work and Energy