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Kinetic Energy and Work. Kinetic Energy. Definition : for a particle moving with speed v, the kinetic energy is. K = ½ mv 2. (a SCALAR quantity). Then the Work-Energy Theorem says:.
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Kinetic Energy and Work Physics 1D03 - Lecture 19
Kinetic Energy Definition: for a particle moving with speed v, the kinetic energy is K = ½ mv2 (a SCALAR quantity) Then the Work-Energy Theorem says: The total work done by all external forces acting on a particle is equal to the increase in its kinetic energy.W = ΔK = Kf – Ki Physics 1D03 - Lecture 19
Kinetic Energy is measured in joules (1J=1N·m). • Kinetic energy is a scalar; the work-energy theorem is a scalar relation. • This theorem is equivalent to Newton’s Second Law. In principle, either method can be used for any problem in particle dynamics. Physics 1D03 - Lecture 19
How to deal with friction If there is friction in the system, then:ΔK=Wf = -ffdSince ΔK = Kf - Ki = -ffdTherefore Kf = Ki - ffd Physics 1D03 - Lecture 19
Example 1 • A bartender slides a 1-kg glass 3 m along the bar to a customer. The glass is moving at 4 m/s when the bartender lets go, and at 2 m/s when the customer catches it. • Find the work done by friction • Calculate the force of friction. Physics 1D03 - Lecture 19
Example 2 • A 6.0 kg block initially at rest is pulled to the right for 3.0m with a force of 12N over a surface. • Determine its final velocity if: • a) the surface has no friction • b) the surface has a coefficient of kinetic friction of 0.15 • How else could we solve this problem ???? • Try it !!! Physics 1D03 - Lecture 19
Solution Physics 1D03 - Lecture 19
Quiz A mass is attached to a horizontal spring and rests on a frictionless table. Starting from the unstretched position at x=0, the spring is displaced by x=A.When does the mass have the highest speed? • when you let it go at x=A • when it goes through x=0 • when it gets to x=-A, on the other side of x=0 • it always has the same speed Physics 1D03 - Lecture 19
Example 3 A block of mass 1.6kg resting on a frictionless surface is attached to a horizontal spring with a spring constant k=1.0x103 N/m (for a spring, E= ½ kx2). The spring is compressed to 2.0cm and released from rest. a) Calculate the speed of the block as it passes the x=0 point.b) Calculate the block’s speed at the x=1.0 cm point. c) Calculate the block’s speed the first time is passes though the x=0cm point if there is a constant frictional force of 4.0 N. Physics 1D03 - Lecture 19
Solution Physics 1D03 - Lecture 19
Example 4 You drop a rock off the top of the CN Tower (h=553.33m). Use the energy-work theorem to determine the rock’s speed as it hits the ground below. Physics 1D03 - Lecture 19
Quiz Your friends at the International Space Station (orbiting at 350 km above the Earth’s surface) were tired of you, and pushed you out of an air lock. Assuming negligible initial speed, if the Earth did not have an atmosphere, how fast would you hit the ground: A) 86 m/s B) 2620 m/sC) 6860 m/sD) depends on the direction you take (straight down, or at an angle) Physics 1D03 - Lecture 19
10 min rest Physics 1D03 - Lecture 19
Potential Energy • Work and potential energy • Conservative and non-conservative forces • Gravitational and elastic potential energy Physics 1D03 - Lecture 19
mg FP = mg y mg y s2 s1 Gravitational Work To lift the block to a height y requires work (by FP :) WP = FPy = mgy When the block is lowered, gravity does work: Wg1 = mg.s1 = mgy or, taking a different route: Wg2 = mg.s2 = mgy Physics 1D03 - Lecture 19
Work done (against gravity) to lift the box is “stored” as gravitational potential energyUg: Ug =(weight) x (height) = mgy (uniform g) When a block moves up, work done by gravity is negative (decrease speed)When a block moves down, work done by gravity is positive (increase speed) • The position where Ug = 0 is arbitrary. • Ug is a function of position only.(It depends only on the relative positions of the earth and the block.) • The work Wg depends only on the initial and final heights, NOT on the path. Physics 1D03 - Lecture 19
Example • A rock of mass 1kg is released from rest from a 10m tall building. What is its speed as it hits the ground ? • The same rock is thrown with a velocity of 10m/s at an angle of 45o above the horizontal. What is its speed as it hits the ground. Physics 1D03 - Lecture 19
Example • What minimum speed does a 100g puck need to make it to the top of a 3.0m long 200 frictionless ramp? Physics 1D03 - Lecture 19
Conservative Forces B path 1 A force is called “conservative” if the work done (in going from A to B) is the same for all paths from A to B. A path 2 W1 = W2 An equivalent definition: For a conservative force, the work done on any closed path is zero. Total work is zero. Physics 1D03 - Lecture 19
Quiz The diagram at right shows a force which varies with position. Is this a conservative force? • Yes. • No. • Maybe, maybe not. Physics 1D03 - Lecture 19
Quiz The diagram at right shows a force which varies with position. Is this a conservative force? • Yes. • No. • Maybe, maybe not. Physics 1D03 - Lecture 19
For every conservative force, we can define a potential energy function U so that WAB=-DU = UA-UB Note the negative Since: W = ∆K = - ∆U : Kinetic ↔ Potential Examples: Gravity (uniform g) : Ug = mgy, where y is height Gravity (exact, for two particles, a distance r apart): Ug=- GMm/r, where M and m are the masses Ideal spring: Us = ½ kx2,where x is the stretch Electrostatic forces (we’ll do this in January) Physics 1D03 - Lecture 19
Non-conservative forces: • friction • drag forces in fluids (e.g., air resistance) Friction forces are always opposite to v (the direction of f changes as v changes). Work done to overcome friction is not stored as potential energy, but converted to thermal energy. Physics 1D03 - Lecture 19
Conservation of mechanical energy If only conservative forces do work, potential energy is converted into kinetic energy or vice versa, leaving the total constant. Define the mechanical energy E as the sum of kinetic and potential energy: E K + U = K + Ug + Us +... Conservative forces only: W = -DU Work-energy theorem: W = DK So, DK+DU = 0; which means that E does not change with time: dE/dt = 0 Physics 1D03 - Lecture 19
Example: Atwood’s Machine An Atwood's machine supports masses m1=0.205 kg and m2=0.292 kg. The masses are held at rest beside each other and then released. Once released the 0.292 kg mass accelerates downward. Neglecting friction, what is the speed of the masses the instant each has moved through 0.424 m? Physics 1D03 - Lecture 19
Example: Pendulum L • The pendulum is released from rest with the string horizontal. • Find the speed at the lowest point (in terms of the length L of the string). vf Physics 1D03 - Lecture 19
Example: Pendulum θ • The pendulum is released from rest at an angle θ to the vertical. • Find the speed at the lowest point (in terms of the length L of the string). vf Physics 1D03 - Lecture 19
10 min rest Physics 1D03 - Lecture 19
Energy Examples Physics 1D03 - Lecture 19
Example • Tarzan (mass 90kg) swings on a 10m long rope off the top of BSB to save a student (mass 60kg) from falling into a construction hole on campus. If Tarzan starts with the rope in a horizontal position and picks up the student at the bottom of the swing, determine how high they will go. Physics 1D03 - Lecture 19
Example • In a ballistic pendulum a bullet is shot into a block on a string. If the block and bullet swing up by a vertical distance of h, determine the speed of the bullet. Physics 1D03 - Lecture 19
Example • A pendulum of length 1m and mass of 100g is released from an angle 30o. At the bottom of the swing hits a spring of spring constant k=10N/m. Determine the maximum compression of the spring. Physics 1D03 - Lecture 19