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Bellringer 10/25

Bellringer 10/25. A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find µ s and µ k between the clock and the floor.

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Bellringer 10/25

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  1. Bellringer 10/25 • A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find µs and µk between the clock and the floor.

  2. Chapter 5: Work & Energy The Work/Energy Relationship

  3. Work done by a constant net force: • W = F∆x • For forces applied at an angle, θ: • W=Fcosθd • “d” is the magnitude of displacement • Work is the use of force to displace an object in the direction of the force. • If an object is not displaced when a force is applied to it, then no work has been done on that object.

  4. Work is only done when a force causes an object to move some distance. • Work is measured in units called Joules (J). • Work is a scalar quantity that can be positive or negative.

  5. Positive work results in a displacement in the same direction as the applied force. • Work is negative when the force is opposite the displacement. Fnet Displacement Friction is doing negative work on the box. Fk Fnet Displacement

  6. Work can be positive or negative • Man does positive work lifting box • Man does negative work lowering box • Gravity does positive work when box lowers • Gravity does negative work when box is raised

  7. Example • How much work is done on a vacuum cleaner pulled 3.0m by a force of 50.0N at an angle of 30 degrees above the horizontal?

  8. Example • A student drags a physics teacher for 25 m on the end of a rope that makes a 40 degree angle with the ground. The force on the rope is 650 N. How much work is done?

  9. Frictional work is done whenever the force of friction hinders motion. • For instance, when pulling a sled over concrete, a significant amount of friction prevents the sled from being pulled as far. Work done by friction: Wf = FkΔx

  10. Applying force at an angle can alter the frictional force. Fy Fy Reduction of frictional force because Fy is upward, and reduces the normal force. Increase of frictional force because Fy is downward, and increases the normal force.

  11. The net work done on an object can be expressed as the change of the object’s kinetic energy. • This expression is known as the “Work - Kinetic Energy Theorem.” • Kinetic Energy is the energy of an object due to its motion. Net Work Done on an Object/ Work-Kinetic Energy Theorem: Wnet = ∆KE = ½mv2 – ½mv02 Kinetic Energy of an Object: KE = ½mv2

  12. Example • A 7.00 kg bowling ball moves at 3.00 m/s. How fast must a 2.45 g table tennis ball move in order to have the same kinetic energy as the bowling ball? Is this speed reasonable for a table tennis ball in play?

  13. Example • On a frozen pond, a person kicks a 10.0kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?

  14. Bellringer 10/31 • Determine the amount of work that must be done by the engine of a 500 kg racing car to change the velocity of the car from 55 ms-1 east to 60 ms-1 east. • If this change in velocity was accomplished in 0.3 s, calculate the acceleration of the car. • Find the net force applied by the engine to cause this acceleration

  15. Review • Work – only done on an object when a net force acts to displace it (W=Fd) • Energy • Objects in motion have kinetic energy (KE = 1/2mv2) • Net work done on or by an object is equal to the change in kinetic energy of the object (W=ΔKE)

  16. Review • How much work is done in lifting a 300N rock 10m off the ground? • Calculate the kinetic energy of a 3.1kg toy cart that moves at 4.8m/s. Calculate the kinetic energy of the same cart at twice the speed. • A car is travelling at 27 m/s north and has a mass of 1500 kg. Calculate the kinetic energy of the car.

  17. Friction is an example of a nonconservative force – one that randomly disperses the energy of the objects on which it acts. • For example, the car shown is undergoing frictional forces as it slides. • The energy is being dissipated as sound waves and thermal energy.

  18. Gravity is an example of a conservative force. It does not dissipate energy. • In order to reach the top of the cliff, the man had to use energy to work against gravity. • This energy used as work is recovered as KE by diving. • Upon reaching the water, his speed gives him kinetic energy equal to the work he used to climb upward.

  19. In general, a force is conservative if the work it does moving an object between two points is the same regardless of the path taken.

  20. Potential energy is the stored energy that results from an object’s position or condition. • It depends only on the beginning and ending points of motion…not the path taken.

  21. Gravitational PE – potential energy stored in the gravitational fields of interacting bodies • product of an objects mass, gravitational acceleration, and height. • The work done by gravity is the negative of the change in gravitational potential energy. Gravitational Potential Energy: PEg= mgh *Can also be expressed as “weight x height” Work done by gravity: Wg = -( PEf – PEi) = -( mghf– mghi)

  22. Gravitational Potential Energy

  23. Example • A spoon is raised 21.0cm above a table. If the spoon and its contents have a mass of 30.0g, what is the gravitational potential energy associated with the spoon at that height relative to the surface of the table?

  24. Bellringer 11/1 • A student slides a 0.75kg textbook across a table, and it comes to rest after traveling 1.2m. Given the coefficient of kinetic friction between the book and the table is 0.34, use the work-kinetic energy theorem to find the book’s initial speed?

  25. Elastic Potential Energy: PEelastic= ½ kx2 k = spring/force constant x = distance compressed/stretched Elastic Potential Energy – energy available when a deformed elastic object returns to its original configuration

  26. Example • A 70.0kg stuntman is attached to a bungee cord with an unstretched length of 15.0m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee is 71.8N/m, what is the total potential energy relative to the water when the man stops falling?

  27. Example continued Given:m= 70.0 kg k = 71.8 N/m g = 9.81 m/s2 h = 50.0 m – 44.0 m = 6.0 m x = 44.0 m – 15.0 m = 29.0 m PE = 0 J at river level Unknown:PEtot= ?

  28. Example • A 2.00kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00m, and the height of the room is 3.00m. What is the gravitational potential energy associated with the ball relative to each of the following? • The ceiling • The floor • A point at the same elevation as the ball

  29. Example • 1. A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? • 2. If a force of 14.7 N is used to drag the loaded cart (from previous question) along the incline for a distance of 0.90 meters, then how much work is done on the loaded cart?

  30. Conservation of Energy • The Law of Conservation of Energy states: ENERGY CAN NEVER BE CREATED OR DESTROYED. • Mechanical Energy (ME) • Sum of kinetic energy and all forms of potential energy • ME = KE + ∑PE

  31. Applying this to KE and PE, we develop the Law of Conservation of Mechanical Energy: • The total mechanical energy of an isolated system will remain constant. Conservation of Mechanical Energy: KEi + PEi = KEf + PEf If gravity is the only force doing work: ½mv2i + mgyi = ½mv2f + mgyf

  32. From the Law of Conservation of Energy, we know that: KEi + PEi = KEf + PEf We can expand this to: ½mvi2 + mghi = ½mvf2 + mghf Suppose an object held at rest at some initial height is dropped, and allowed to impact the ground. Rearrange the above equation to solve for the final velocity of the object (at the moment when it impacts).

  33. Example • Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0kg.

  34. Example • You are designing a roller coaster in which a car will be pulled to the top of a hill of height h and then, starting from a momentary rest, will be released to roll freely down the hill and toward the peak of the next hill, which is 1.1 times as high. Will your design be successful? Explain your answer.

  35. Explain how work and energy are related (aside from having the same unit). • Use the picture to describe the changes in energy that take place between each point on the coaster. • Which point on the coaster will have the highest KE? Explain.

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