Chapter 7 Work and Kinetic Energy

Chapter 7 Work and Kinetic Energy

Reading and Review

So, as long as: vertical (down) at C: Centripetal acceleration must be C at the top, then N>0 and pointing down. (now apparent weight is in the opposite direction to true weight!) B horizontal vertical (up) A Vertical circular motion Condition for falling: N=0

A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? c e b a d Barrel of Fun

A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? c e b a d Barrel of Fun The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down?

Around the Curve a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn, and you find yourself hitting the passenger door. What is the correct description of what is actually happening?

Around the Curve a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn, and you find yourself hitting the passenger door. What is the correct description of what is actually happening? The passenger has the tendency to continue moving in a straight line. There is a net centripetal force, provided by the door, that forces the passenger into a circular path.

Working Hard... or Hardly Working Sisyphus pushes his rock up a hill. Atlas holds up the world. (b) Who does more work? (a)

Working Hard... or Hardly Working Sisyphus pushes his rock up a hill. Atlas holds up the world. (b) With no displacement, Atlas does no work (a)

Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: SI unit: newton-meter (N·m) = joule, J

Friction and Work I a) friction does no work at all b) friction does negative work c) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

Displacement N Pull f mg Friction and Work I a) friction does no work at all b) friction does negative work c) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction? Friction acts in the opposite direction to the displacement, so the work is negative. Or using the definition of work (W = F (Δr)cos  ), because= 180º, thenW <0.

Friction and Work II Can friction ever do positive work? a) yes b) no

Friction and Work II Can friction ever do positive work? a) yes b) no Consider the case of a box on the back of a pickup truck. If the box moves along with the truck, then it is actually the force of friction that is making the box move.

Forces not along displacement If the force is at an angle to the displacement:

Convenient notation: the dot product The work can also be written as the dot product of the force and the displacement:

Force and displacement The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:

Sum of work by forces = work by sum of forces If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force:

Units of Work 1 kcal = 1 Cal = 4.186 kJ Lifting 0.5 L H2O up 20 cm = 1 J

Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? a) catcher has done positive work b) catcher has done negative work c) catcher has done zero work

Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? a) catcher has done positive work b) catcher has done negative work c) catcher has done zero work The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because = 180º, thenW < 0. Note that the work done on the ball is negative, and its speed decreases. Follow-up: What about the work done by the ball on the catcher?

Tension and Work a) tension does no work at all b) tension does negative work c) tension does positive work A ball tied to a string is being whirled around in a circle. What can you say about the work done by tension?

T v Tension and Work a) tension does no work at all b) tension does negative work c) tension does positive work A ball tied to a string is being whirled around in a circle. What can you say about the work done by tension? No work is done because the force acts in a perpendicular direction to the displacement. Or using the definition of work (W = F (Δr)cos  ), because  =90º, thenW = 0. Follow-up: Is there a force in the direction of the velocity?

N A ball of mass m rolls down a ramp of height h at an angle of 45o. What is the total work done on the ball by gravity? Fgx = Fg sinθ W = Fd = Fgxx L = (Fg sinθ) (h / sinθ) W = Fd = Fg x h a W = mgh h = L sinθ Fg a W = Fg h = mgh Fg h h θ Work by gravity A ball of mass m drops a distance h. What is the total work done on the ball by gravity? Path doesn’t matter when asking “how much work did gravity do?” Only the change in height!

Path independence • If a force depends on POSITION only then the work done by it on an object moving from to will NOT depend upon the path. • Such a force is called a Conservative Force

Motion and energy When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

Kinetic Energy After algebraic manipulations of the equations of motion, we find: Therefore, we define the kinetic energy:

Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (True for rigid bodies that remain intact)

FHAND ∆r v = const a = 0 mg Lifting a Book You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book? a) mg×∆r b) FHAND×∆r c) [FHAND + mg] ×∆r d) zero e) none of the above

FHAND ∆r v = const a = 0 mg Lifting a Book You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book? a) mg×∆r b) FHAND×∆r c) (FHAND + mg) ×∆r d) zero e) none of the above The total work is zero because the net force acting on the book is zero. The work done by the hand is positive, and the work done by gravity is negative. The sum of the two is zero. Note that the kinetic energy of the book does not change either! Follow-up: What would happen if FHAND were greater than mg?

Kinetic Energy I By what factor does the kinetic energy of a car change when its speed is tripled? a) no change at all b) factor of 3 c) factor of 6 d) factor of 9 e) factor of 12

Kinetic Energy I By what factor does the kinetic energy of a car change when its speed is tripled? a) no change at all b) factor of 3 c) factor of 6 d) factor of 9 e) factor of 12 Because the kinetic energy is mv2, if the speed increases by a factor of 3, then the KE will increase by a factor of 9.

Slowing Down a) 20 m b) 30 m c) 40 m d) 60 m e) 80 m If a car traveling 60 km/hr can brake to a stop within 20 m, what is its stopping distance if it is traveling 120 km/hr? Assume that the braking force is the same in both cases.

Slowing Down a) 20 m b) 30 m c) 40 m d) 60 m e) 80 m If a car traveling 60 km/hr can brake to a stop within 20 m, what is its stopping distance if it is traveling 120 km/hr? Assume that the braking force is the same in both cases. F d = Wnet = ∆KE = 0 – mv2, and thus,|F| d = mv2. Therefore, if the speeddoubles, the stopping distance getsfourtimes larger.

Condition:Fcp > mg at top of loop Fcp = mv2/r = mg v2 = gr KE = mv2 / 2 = mgr/2 Gravity must provide this energy Wg = mg∆h = KE ∆h = r/2 above the top of the loop! Application: ball on a track how high must I place the ball so that it can complete a loop?

Work Done by a Variable Force If the force is constant, we can interpret the work done graphically:

Work Done by a Variable Force If the force takes on several successive constant values:

Work Done by a Variable Force We can then approximate a continuously varying force by a succession of constant values.

Work Done by a Variable Force The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is

Hooke’s Law:F = - kx Loaded spring: W = kx2/2= (760 N/m) (0.04m)2/ 2 KE = mv2/2 = (1kg)(1m/s)2 / 2 KE = 0.55 J Kinetic Energy: k = (3kg)(9.8 m/s2) / (3.9 cm) k = 760 N/m W = 0.61 J Application: work by a spring How fast?: v = d/t = (0.020 m) (0.020 s) = 1 m/s

Power Power is a measure of the rate at which work is done: SI unit: J/s = watt, W 1 horsepower = 1 hp = 746 W

Power

Power If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written: Question: what is the total work per unit time done on the object?

Electric Bill a)energy b) power c) current d) voltage e) none of the above When you pay the electric company by the kilowatt-hour, what are you actually paying for?

Electric Bill a)energy b) power c) current d) voltage e) none of the above When you pay the electric company by the kilowatt-hour, what are you actually paying for? We have defined: Power = energy/ time So we see that:Energy = power× time This means that the unit ofpower× time (watt-hour) is a unit ofenergy !!

A block rests on a horizontal frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an angle above the horizontal, as shown in the figure. After the block is pulled through a distance of 1.50 m, its speed is 2.60 m/s, and 50.0 J of work has been done on it. (a) What is the angle (b) What is the mass of the block?

The pulley system shown is used to lift a 52 kg crate. Note that one chain connects the upper pulley to the ceiling and a second chain connects the lower pulley to the crate. Assuming the masses of the chains, pulleys, and ropes are negligible, determine (a) the force F required to lift the crate with constant speed, and (b) the tension in two chains

(a) the force F required to lift the crate with constant speed, and (b) the tension in two chains (a) constant velocity, a=0, so net force =0. 2T - (52kg)(9.8m/s2) = 0 T = 250 N F = -250 Ny (b) upper pulley doesn’t move: Tch - 2Trope = 0 Tch = 500 N lower pulley has constant acceleration Tch -2Trope =0 Tch = 500 N Mechanical Advantage!

What about work? (a) how much power is applied to the box by the chain? (b) how much power is applied on the rope by the applied force? Trope = 250 N Tchain = 500 N F = -250 Ny • (a) P = Fv = 500 N * vbox • P = Fv = 250 N * vhand • hand moves twice as fast • hand moves twice as far

Chapter 7 Work and Kinetic Energy