Work and Energy Physics Mr. Day

Work F Work - the product of the magnitudes of the component of a force along the direction of displacement and the displacement F W = F d F Work –Push a chair from rest to a velocity F Not Work –Hold a book in the air –Carry a chair across the room at a constant velocity

Work (cont.) F Work is only done when the components of a force are parallel to a displacement d F F d W = F d All of the force is acting on the box W = F d cos  Only the horizontal component of the force does work

Sign Convention for Work Visual Concept

Work (cont.) Examples d F d F d F W = F d cos  W = F d cos 0 W= F d W = F d cos  W = F d cos 90 W = 0 dF W = F d cos 

Work (cont.) F Units –F d = W –N m = Joule (J) F Work is a scalar quantity –Work can be positive or negative Positive Work - Force and displacement are in the same directionPositive Work - Force and displacement are in the same direction Negative Work - Force and displacement are in opposite directionsNegative Work - Force and displacement are in opposite directions

Work Defined Video

EX: A 20.0 kg suitcase is raised 3.0 m above a platform by a conveyor belt. How much work is done on the suitcase m = 20.0 kg d = 3.0 m W = F d W = (m a) d W = (20.0 kg)(9.8 m/s 2 )(3.0 m) W = 588 J

EX: A person drags a suitcase with a N force at an angle of 60.0° for m. How much work does she do? F = N  = 60.0° d = m d F W = F d cos  W= (100.0 N)(200.0 m) cos 60° W = 10,000 J W = 1.0 X 10 4 J

Types of Energy Video

Energy F Kinetic Energy - the energy of an object due to its motion –Depends on speed and mass –KE = 1/2 m v 2 –Units  Joules (J) F Potential Energy - energy associated with an object due to its position –Units  Joules (J)

Kinetic and Potential Energy Video

Kinetic Energy Visual Concept

EX: A 6.0 kg cat runs after a mouse at 10.0 m/s. What is the cat’s kinetic energy? m c = 6.0 kg v c = 10.0 m/s KE c = 1/2 m c v c 2 KE c = 1/2 (6.0 kg)(10.0 m/s) 2 KE c = 300 J

EX: If a.10 kg mouse runs as fast as the cat, what is its kinetic energy? m m =.10 kg v m = 10.0 m/s KE m = 1/2 m m v m 2 KE m = 1/2 (.10 kg)(10.0 m/s) 2 KE m = 5 J

EX: A 24 kg dog begins to chase the cat and has the same kinetic energy as the cat. What is the dog’s velocity? m d = 24 kg KE d = 300 J KE d = 1/2 m d v d 2 v d = √(2 Ke d / m d ) v d = √(2 (300 J) / 24 kg) v d = 5.0 m/s

Potential Energy F Gravitational Potential Energy - potential energy associated with an object due to its position relative to Earth or some other gravitational source PE g = m g h F Elastic Potential Energy - the potential energy in a stretched or compressed elastic object –Spring –Rubber band

Potential Energy Visual Concept

Elastic Potential Energy F The length of a spring when no external forces are acting on it is called the relaxed length F PE e = 1/2 k x 2 –PE e = 1/2 (spring constant)(distance stretched or compressed) 2 F Spring constant - a parameter that expresses how resistant a spring is to being compressed or stretched

Elastic Potential Energy

Spring Constant F High spring constant  stiff spring F Low spring constant  flexible spring F k = F / d –Units  N / m

Spring Constant Visual Concept

EX: When a 2.00 kg mass is attached to a vertical spring, the spring is stretched 10.0 cm so the mass is 50.0 cm above the table. m = 2.00 kg x = 10. cm =.10 m h = 50.0 cm =.50 m

A. What is the gravitational potential energy associated with the mass relative to the table? PE g = mgh PE g = (2.00 kg)(9.8 m/s 2 )(.50 m) m = 2.00 kg x = 10. cm =.10 m h = 50.0 cm =.50 m PE g = 9.8 J

B. What is the elastic potential energy if the spring constant is N/m? PE e = 1/2 k x 2 PE e = 1/2 (400.0 N/m)(.10m) 2 m = 2.00 kg x = 10. cm =.10 m h = 50.0 cm =.50 m k = N/m PE e = 2.00 J

C. What is the total potential energy of the system? ∑PE = PE g + PE e ∑PE = 9.8 J J m = 2.00 kg x = 10. cm =.10 m h = 50.0 cm =.50 m k = N/m PE e = 11.8 J

Mechanical Energy F There are many types of energy associated with a system –Kinetic –Gravitational potential –Elastic potential –Chemical –Thermal Most can be ignored because they are negligible or not relevantMost can be ignored because they are negligible or not relevant

Mechanical Energy F Mechanical energy - the sum of the kinetic and all forms of potential energy F ME = ∑KE + ∑PE F All other forms of energy are classified as non-mechanical energy

Conservation of Energy F Conserve means it remains the same F Conservation of mechanical energy ME i = ME f ∑KE i + ∑PE gi + ∑PE ei = ∑KE f + ∑PE gf + ∑PE ef F In the presence of friction, energy is “lost” to heat energy

Niagara Falls and Energy Transformation Video

Conservation of Mechanical Energy Visual Concept

F As the roller coaster falls the energy is transformed from potential energy to kinetic energy F The energy is then transferred back into potential energy, etc. Energy of a roller coaster

Energy is a sling shot F It starts with elastic potential energy F It quickly transfers into kinetic energy F As the height increases it transfers into gravitational energy F As it falls the energy transfers into kinetic energy

EX: A small 10.0 g ball is held to a slingshot that is stretched 6.0 cm. The spring constant of the band on the slingshot is 2.0 X 10 2 N/m. A. What is the elastic potential energy of the slingshot before it is released PE e = 1/2 k x 2 PE e = 1/2 (2.0 X 10 2 N/m)(.06 m) 2 m = 10.0 g =.0100 kg x = 6.0 cm =.06 m k = 2.0 X 10 2 N/m PE e =.36 J

B. What is the kinetic energy of the ball just after the slingshot is released? ME i = ME f.36 J = ∑KE f ∑KE i + ∑PE gi + ∑PE ei = ∑KE f + ∑PE gf + ∑PE ef ∑PE ei = ∑KE f

C. What is the balls speed at the instant it leaves the slingshot? KE f = 1/2 m v 2 v = √ (2KE f / m) v = √(2(.36 J) / (.01 kg)) v = 8.5 m/s

D. How high would the ball travel if it were shot directly upward? ME i = ME f.36 J = ∑PE f ∑KE i + ∑PE gi + ∑PE ei = ∑KE f + ∑PE gf + ∑PE ef ∑KE i = ∑PE gf PE f = mgh h = PE f / mg h =.36 J / ((.01 kg)(9.8 m/s 2 )) h = 3.7 m

Work and Energy F Work-kinetic energy theorem - the net work done on an object is equal to the change in the kinetic energy of the object F W net = ∆ KE F W friction = ∆ ME

Work Kinetic Energy Theorem Visual Concept

F Potential Energy is transferred into Kinetic Energy F Next the change in the Kinetic Energy is equal to the net work

Stopping Distance F If an object has a higher kinetic energy, more work is required to stop the object

Ex: On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does it travel if the coefficient of kinetic friction between the sled and the ice is.10? m = 10.0 kg v i = 2.2 m/s v f = 0 m/s µ k =.10 vivi W = ∆KE F d = KE f - KE i µ k = F k / F N F k = µ k (-mg) µ k (-mg) d = 1/2 m v f 2 - 1/2 m v i 2 µ k (-mg) d = - 1/2 m v i 2 d = (- 1/2 m v i 2 ) / µ k (-mg) d = (- 1/2 (10.0 kg) (2.2 m/s) 2 ) / (.10 (-10.0 kg) (9.8 m/s 2 )) d = 2.47 m

EX: A 10.0 kg shopping cart is pushed from rest by a N force against a 50.0 N friction force over 10.0 m distance. m = 10.0 kg v i = 0 F p = N F k = 50.0 N d = 10.0 m FpFp FkFk FNFN FgFg

A. How much work is done by each force on the cart? W p = F p d cos  W p = (250.0 N)(10.0 m)cos 0 W p = 2500 J W k = F k d cos  W k = (50.0 N)(10.0 m)cos 180 W k = -500 J W g = 0 W N = 0

B. How much kinetic energy has the cart gained? W net = ∆KE W p + W k = KE f - KE i 2500 J J = KE f - 0 KE f = 2000J C. What is the carts final speed? KE = 1/2 m v 2 v = √((2KE)/(m)) v = √((2(2000 J))/(10.0 kg)) v = 20 m/s

Power F Power - the rate at which energy is transferred F P = W / ∆ t –Units  J / s  watt (w) F Since P = W / ∆ t and W = F d; –P = F d / ∆ t  P = F (d / t) –P = F v F Horsepower is another unit of power –1 hp = 746 w

Power Defined Video

Power Visual Concept

EX: A N force moves an object 20.0 m in 5.0 s. What is the power? F = N d = 20.0 m d = 20.0 m t = 5.0 s P = 400 w P = F (d / t) P = N (20.0 m / 5.0 s) d = 20.0 m F = N

EX: Two horses pull a cart. Each exerts a 250 N force at a 2.0 m/s speed for 10.0 min. F h1 = 250 N F h2 = 250 N v = 2.0 m/s ∆ t = 10.0 min = 600 s v = 2.0 m/s F = 250 N

A. Calculate the power delivered by the forces. P h1 = F h1 v P h1 = (250 N)(2.0 m/s) P h1 = 500 w P h2 = F h2 v P h2 = (250 N)(2.0 m/s) P h2 = 500 w ∑P = P h1 + P h2 ∑P = 500 w + 500w ∑P = 1000 w

B. How much work is done by the two horses? P = W / ∆ t W = P ∆ t W = (1000 w)(600 s) W = 6.0 X 10 5 J

Work Cited F Sources – – –Holt Physics –United Streaming