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Work and Kinetic Energy

Work and Kinetic Energy. Work. In Physical science, work is the use of force to move an object some distance. In scientific terms, you do work only when you exert on force on an object and move it. Work is done on an object only if the object moves in the same direction as the force. .

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Work and Kinetic Energy

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  1. Work and Kinetic Energy

  2. Work • In Physical science, work is the use of force to move an object some distance. In scientific terms, you do work only when you exert on force on an object and move it. • Work is done on an object only if the object moves in the same direction as the force.

  3. Calculating Work • You can calculate the work a force does if you know the size of the force applied to an objects and the distance over which the force acts. The distance involved is the distance the object moved in the direction of that force. The calculation for work is shown in the following formula: Work = force . distance W = Fd

  4. Example: How much work does the force of gravity do when 25 N object falls a distance of 3.5 m? W = Fd = (25 N) (3.5m) W = 87.5 J

  5. If a force is exerted perpendicular to the motion, no work is done. What if a force is exerted at some other angle to the motion? • The magnitude of the component of the force F acting in the direction of motion is found by multiplying the force F by the cosine of the angle between F and the direction of motion. W = F(coso)d = Fd coso

  6. Example: A sailor pulls a boat along a dock using a rope at an angle of 70o with the horizontal. How much work is done by the sailor if he exerts a force of 240 N on the rope and pulls the boat 30.0 m? W = Fd coso = (240N)(30.0m)(cos70o) = (240N)(30.0m)(0.63) W= 4536 J

  7. Work done against gravity can be found using the equation Work = Force . height or W = Fh. Since F = mg, we can use the equation W = mgh ( where m = mass, g = gravity, or 9.8 m/s² and h = height). Ex: You lift a 2 kg book and put it on a shelf 3 meters high. How much work did you do? W = mgh = (2kg)(9.8 m/s²)( 3m) W= 58.8 J

  8. Work Transfers Energy • Energy is the ability of a person or an object to do work or to cause a change. When you do work on an object, some of your energy is transferred to the object. You can think of work as the transfer of energy. In fact, both work and energy are measured in the same unit, the joule. If you put energy into an object, then you do work on that object.

  9. Kinetic Energy • Is a form of energy that moving objects have. • The energy of motion.

  10. Forms of kinetic energy • Vibrational Energy: It is the energy due to Vibrational motion. • Rotational Energy: It is the energy due to rotational motion.  • Translational Energy: It is the energy due to motion from one location to another. • Relativistic energy: Itis the energy of objects operating at very high speed.

  11. Calculating Kinetic Energy • Kinetic energy can be calculated using the following formula: Kinetic Energy = mass . velocity2 2 KE = ½ mv2

  12. If you double the mass of an object, you double its kinetic energy. • If you double the object’s velocity, its kinetic energy is four times greater.

  13. Here’s a question not many ask: why is the velocity squared in the kinetic energy equation, E = ½mv2. Why should the energy depend on the square of the velocity? To find out, let us look at how the first equation is derived: • We start with the constant acceleration equation, 2ad = vf2 - vi2 • Then substitute a = F/m into that 2Fd/m = vf2 - vi2 • If we let the initial velocity equal zero, and define work as force through a distance, we get W = E = Fd = ½ mvf2

  14. Not all objects start at rest. They may already have kinetic energy when additional work is done on them. If we define initial kinetic energy, Kei, and final kinetic energy, Kef, as the energies the object has before and after work is done, we can write W = AKE or W= KEf - KEi • This equation represents the work energy theorem. It can be stated: The net work done on an object is equal to its change in kinetic energy.

  15. Example: A shot putter heaves a 7.26 kg shot with final velocity of 7.50 m/s. • What is the kinetic energy of the shot? • The shot was initially at rest. How much work was done on it to give it this kinetic energy? Solution: a) KE = ½ mv2 = ½(7.26 kg)(7.50m/s)2 = 204 kg.m2/s2 KE = 204 J b) Work done is equal to change in the kinetic energy. W = KEf - KEi = 204 J – 0 J W = 204 J

  16. Reporters: PELIN, Kaina Angelique PAQUEO, Caedmon Allan MOSENDE, Kristina Libra CALIAO, Lyeziel MAKINANO, Clementine JORTA, Myrine III - NEWTON

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