Kinetic Energy and Work AP Physics Chapter 7 Kinetic Energy and Work

AP Physics Turn in Chapter 6 Homework, Worksheet & Lab Report Lecture Q&A

Kinetic Energy, K Kinetic Energy: ability to do work due to motion m: mass of object v: velocity of object Unit: [K] = [m] · [v]2 = kg · (m/s)2 = Joule = J 1 electron-volt = 1 eV = 1.60  10-19 J Scalar, no direction K  0

Example: What is the kinetic energy associated with the Earth’s orbiting of the Sun? (Mass of Earth is 5.98  1024 kg, and radius of Earth orbiting around sun is 1.50  1011 m.) m = 5.98  1024 kg, T = 365 dys = 3.15  107 s r = 1.50  1011 m, K = ?

Work, W Work: energy transferred to or from an object by means of a force acting on the object Work done by object A (to object B) W > 0 if energy transferred (from A) to object B W < 0 if energy transferred from object B (to A)

Work Done by A Constant Force  Work Done by A Constant Force d W: work done by force on the object F: constant force acting on object d: displacement of object with force acting on, x : angle between F and d  < 90o  W > 0  > 90o  W < 0  = 90o  W = 0 W: a scalar, no direction (but W can be negative) Valid for particle-like objects

Work Done by Constant Forces Find the net force (vectors), then find the dot product of the net force and the displacement, or Find the work done by each individual force, then add up all the works done by each force.

Practice: Pg160-8 A floating ice block is pushed through a displacement d = (15 m)i – (12 m) j along a straight embankment by rushing water, which exerts a force F = (210 N)i – (150N )j on the block. How much work does the force do on the block during the displacement?

Work-Kinetic Energy Theorem The total work done on an object is equal to the change in the kinetic energy of that object Another look at work: If F and d are in same direction:

DE: K: __  __ (though v: __  __)  WDE __ 0 + - > Practice: A single force acts on a body that moves along a straight line. The diagram shows the velocity versus time for the body. Find the sign (plus or minus) of the work done by the force on the body in each of the intervals AB, BC, CD, and DE. t v A C B E D + - W = K AB: K: __ __  K __ 0  WAB __ 0 + > > BC: v __ 0  K __ 0  WBC __ 0 = = = CD: K: __  __  K __ 0  WCD __ 0 + < < DE: K: __  __ (though v: __  __)  WDE __ 0 + - >

Practice: Pg160-13 Figure 7-29 shows three forces applied to a greased trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F1 = 5.00 N, F2 = 9.00 N, and F3 = 3.00 N. During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease? 60o F2 F3 F1

Work (Wg) Done by Weight (Fg) m: mass g = 9.8 m/s2 d = displacement : angle between d and downward direction h 180o mg h: change in object’s height ? Going up  h > 0  Wg < 0  Weight doing negative work Going down  h __ 0  Wg ___ 0  Weight doing ______ work < > positive

Practice: Pg164-59 To push a 25.0 kg crate up a frictionless incline, angled at 25.0o to the horizontal, a worker exerts a force of 209 N, parallel to the incline. As the crate slides 1.50 m, a) How much work is done on the crate by the worker’s applied force, b) How much work is done on the crate by the weight of the crate, and c) How much work is done on the crate by the normal force exerted by the incline on the crate? d) What is the total, work done on the crate?  h d

Practice: Pg161-19 A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. when the block has been lowered a distance d, find a) the work done by the cord’s force on the block, b) the work done by the weight of the block, c) the kinetic energy of the block, and d) the speed of the block.

Solution: Pg 161-19 Let downward be positive direction, then T + W

Work Done by a Variable Force or: Area under curve of Force-position graph

Example: Pg162-31 A 10 kg mass moves along an x axis Example: Pg162-31 A 10 kg mass moves along an x axis. Its acceleration as a function of its position is shown in Fig. 7-37. What is the net work performed on the mass by the force causing the acceleration as the mass moves from x = 0 to x = 8.0 m?

Spring Force When a spring is compressed or stretched, it exerts a force on the object that compresses or stretches it. F x Force is opposite to change of length of spring F x  Restoring force

Hooke’s Law Spring force is given by F: Force exerts by spring, not by you k: Spring constant, property of spring, stiffness of spring (larger k, stiffer spring) x: Displacement of (end of) spring from equilibrium position, x -: F and x are in opposite direction Notice that this force is the force exerted by the spring, not the force you apply to compress or stretch the spring.

Ideal (Perfect) Spring No mass Can always go back to original length when not compressed or stretched. Does not become warm/hot x is small compared to the entire length of spring

Work Done by Spring (Force) This is the work done by the spring, not the work done (by you) to compress or stretch the spring. If the spring is initially at equilibrium position, then xi = 0, and - work ? F F x x Xi = 0 xf xf

Example: A spring with a spring constant of 15 N/cm has a cage attached to one end. How much work does the spring force do on the cage if the spring is stretched from its relaxed length by 7.6 mm? How much additional work is done by the spring force if it is stretched by an additional 7.6 mm? equilibrium 7.6 mm 7.6 mm

Practice: The only force acting on a 2.0 kg body as it moves along the positive x axis has an x component Fx = -6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s. What is the velocity of the body at x = 4.0 m? At what positive value of x will the body have a velocity of 5.0 m/s?

Solution

Power Power: the (time) rate at which work is done by a force, or the (time) rate at which energy is transferred (from one object to another) Average Power: W: work (scalar, no direction) t: time, duration, for that amount of work done Unit of Power: 1 horsepower = 1 hp = 746 W

Instantaneous Power F: force, vector v: velocity, vector : angle between F and v

Example: Pg165-73 An elevator cab has a mass of 4500 kg and can carry a maximum load of 1800 kg. If the cab is moving upward at full load of 3.80 m/s, what power is required of the force moving the cab to maintain that speed?

Practice: Pg162-44 At a certain instant, a particle experiences a force F = (4.0 N)i – (2.0 N)j + (9.0 N)k while having a velocity v = -(2.0 m/s)i + (4.0 m/s)k. What is the instantaneous rate at which the force does work on the particle? At some other time, the velocity consists of only a j component. If the force is unchanged, and the instantaneous power is –12 W, what is the velocity of the particle just then?

Solution: Pg162-44 b) v = ? Let the velocity be v = vy j, then So, v = (6.0m/s) j

Practice: Pg162-45 A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min. The elevator’s counterweight has a mass of only 950 kg. so, the elevator motor must help pull the cab upward. What average power (in horsepower) is required of the force that motor exerts on the cab via the cable?

Solution: Pg162-45 Three things doing work: Work done by gravity on elevator cab: We = -meg h = -1200kg  9.8m/s2  54 m = -6.35  105 J Work done by gravity on counterweight: Wc = -mcgh = -950kg  9.8m/s2  (-54 m) = 5.03  105 J Work done by motor: Wm = ? Constant velocity: W = 0  Wm + We + Wc = 0 So, Wm = -(We + Wc ) = - (-6.35  105 J + 5.03  105 J) = 1.32  105 J Then average power is

Another approach The motor essentially needs to do work to raise the cab (1200 kg) 54 m and lower the counterweight (950kg) 54 m. So equivalently, the motor needs to do work to raise a mass of 250 kg (1200kg – 950kg) 54 m. The force needed is So the work done by the motor is And the power is

Reference Frames Invariant quantities: quantities measured exactly the same at different inertial reference frames force, mass, acceleration, time Variant quantities: displacement, velocity Principle of Invariance: The laws of physics must have the same form in all inertial reference frames.