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Lecture 19 Gravitational Potential Energy; Conservation of Energy; Conservative and Non- Conservative Forces

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Text of Lecture 19 Gravitational Potential Energy; Conservation of Energy; Conservative and Non-...

  • Lecture 19: Gravita/onal Poten/al Energy; Conserva/on of Mechanical Energy;

    Conserva/ve and non- conserva/ve forces

  • Lecture Objectives 1. Define the gravitational potential energy in terms of

    the work done on an object by a uniform gravitational force.

    2. Apply the conservation of mechanical energy to analyze motion of an object (or a system of objects) subject to uniform gravitational force.

    3. Differentiate conservative from non-conservative forces.

    4. Calculate the work done by a non-conservative force.

  • The property of an object or system that enables it to do work is energy. Like work, energy is measured in joules.

    Mechanical energy is the energy due to the position or movement of an object.

    Mechanical energy

    Kinetic energy Potential energyIn motion: K or KE Position: U or PE

  • Potential energy is stored energy.

    4

    Potential energy is associated with position.

    Potential energy is the measure of the change in the state of motion of the particle

    There are many types of potential energy; - due to Earths gravity - elasticity of a spring

  • Gravitational potential energy is associated with a bodys weight and height above the ground.

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  • Gravitational potential energy is energy stored by raising a mass from the Earth.

    6

  • Example You lift a 100-N boulder 1 m. 1. How much work is done on the boulder? 2. What power is expended if you lift the boulder in a time of 2 s? 3. What is the gravitational potential energy of the boulder in the lifted position?

    Answer: 1. W = Fd = 100 Nm = 100 J 2. Power = 100 J / 2 s = 50 W 3. Relative to its starting position, the boulders PE is PE = mgh = (100N)(1m) =100 J.

  • What is the potential energy? a. The boulder is lifted with 100 N of force. b. The boulder is pushed up the 4-m incline with 50 N

    of force. c. The boulder is lifted with 100 N of force up each 0.5-

    m stair.

  • Conservation of Mechanical Energy

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    Energy cannot be created nor destroyed but converted another form

    http://www.physicsclassroom.com/mmedia/energy/ie.cfm

  • Conservation of Mechanical Energy

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  • When the woman leaps from the burning building, the sum of her PE and KE remains constant at each successive position all the way down to the ground.

    Conservation of Mechanical Energy

  • Problem Solving: Conservation of Energy1. Identify initial and final point (1 or 2)2. Set-up the (total) initial and final kinetic (K1 and K2) and potential energies (U1 and U2) 3. Set-up energy conservation (E1 = E2) 4. Solve for the unknown

  • Given: m = 0.145kg v = 20.0m/s Required: h

    Sample Problem: Young and Freedman You throw a 0.145-kg baseball straight up in the air, giving it an initial upward velocity of magnitude 20.0m/s. Find how high it goes, ignoring air resistance.

  • To solve the problem: calculate the mechanical energy at point 1 and 2

  • Calculate the mechanical energy at point 1 and 2

    At point 1: m = 0.145kg y1 = 0 v1 = 20.0m/s

    E1 = U1 + K1 E1 = mgy1 + mv12

    E1 = 0 + (0.145kg)(20m/s)2

  • Calculate the mechanical energy at point 1 and 2

    At point 2: m = 0.145kg y2 = ??? v2 = 0

    E2 = U2 + K2 E2 = mgy2 + mv22

    E2 = (0.145kg)(9.8m/s2)(y2) + 0

  • Conservation of energy says that: E1 = E2

    Therefore; (0.145kg)(20m/s)2 = (0.145kg)(9.8m/s2)(y2)

    Do the math and solve for y2:

    y2 = 20.04m

  • 18

    Sample Problem: Young and Freedman Throcky skateboards down a curved playground ramp. If we treat Throcky and his skateboard as a particle, he moves through a quarter-circle with radius R = 3.00m. The total mass of Throcky and his skateboard is 25kg. He starts from at rest and there is no friction. (a) Find his speed at the bottom of the ramp. (b) Find the normal force that acts on him at the bottom of the curve.

    Given: R = y1 = 3.00m y2 = 0 v1 = 0 m = 25kg Required: v2 & n2

  • The initial and final kinetic and potential energies are:

    Given: R = y1 = 3.00m y2 = 0 v1 = 0 m = 25kg Required: v2 & n2

  • To solve for n2: calculate for the radial acceleration at point 2:

    Using the free-body diagram at point 2, we can calculate n2 as:

    From conservation of mechanical energy:

  • Conservative forces

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    Conservative forces offer the opportunity of direct conversion between kinetic and potential energies.

    The work done by this force is zero if the particle returns to the initial position.

    The work done by these forces is independent of the path.

    When the only forces that do work in a system are conservative, the total mechanical energy of the system

    is conserved : Ei = Ef

  • Non-conservative forces

    22

    Non-conservative forces are referred to as dissipative forces (cause mechanical energy to be lost or dissipated.

    Work done by non-conservative forces cannot be represented by a potential energy function.

    The process is irreversible.

    In non-conservative forces are present in the system, the

    total mechanical energy is not conserved. Ei > Ef

  • Conservative forces Examples: gravitational force, spring force

    Non-conservative force Examples: friction, fluid resistance

  • Seatwork

  • Seatwork Problem: Young and Freedman A baseball is thrown from the roof of a 22.0 m tall building with an initial velocity of magnitude 12.0 m/s.

    Seatwork Questions: 2) what is the velocity of the baseball just before it hits the ground? 3) what is the velocity of the baseball just before it hits the ground if the initial velocity is aimed at an angle 53.1o below the horizontal

  • Seatwork Problem: Young and Freedman A baseball is thrown from the roof of a 22.0 m tall building with an initial velocity of magnitude 12.0 m/s.

    Seatwork Questions: 2) what is the velocity of the baseball just before it hits the ground? 3) what is the velocity of the baseball just before it hits the ground if the initial velocity is aimed at an angle 53.1o below the horizontal

  • At point 1: y1 = 22.0m v1 = 12.0m/s

    E1 = U1 + K1 E1 = mgy1 + mv12

    E1 = m(9.8m/s2)(22.0m) + m(12.0m/s)2

    At point 2: y2 = 0 v2 = ???

    E2 = U2 + K2 E2 = mgy2 + mv22

    E2 = 0 + mv22

    E2 = mv22

  • Conservation of energy says: E1 = E2

    Therefore: m(9.8m/s2)(22.0m) + m(12.0m/s)2 = mv22

    (Notice that m will be cancelled here) Do the math and solve for v2 (2) v2 = 24.0m/s

    Since the angle did not matter (3) v2 = 24.0m/s

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