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Energy

Energy. General Physics. Kinetic Energy. Energy a moving object has K = ½ mv 2 m: mass (kg) V: velocity (m/s) K: kinetic energy (joules). Work-Energy Theorem. D K = W or D PE = W D K = K f – K i D PE = PE f - PE i. Potential Energy.

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Energy

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  1. Energy General Physics

  2. Kinetic Energy • Energy a moving object has • K = ½ mv2 • m: mass (kg) • V: velocity (m/s) • K: kinetic energy (joules)

  3. Work-Energy Theorem DK = W or DPE = W DK = Kf – Ki DPE = PEf - PEi

  4. Potential Energy • Stored energy due to gravitational attraction between Earth and all objects • PE = mgh • M: mass (kg) • g: 9.8 m/s2 • h: height object is from reference point (meters) • PE:potential energy (joules)

  5. Conservation of Energy • Energy can not be created nor destroyed, it can only change forms

  6. Mechanical Energy • Energy acquired by objects upon which work is done • ME= PE + K

  7. Example Problem • A 15.0 kg cart is moving with a velocity of 7.50 m/s down a hallway. A constant force of 10.0 N acts on the cart, and its velocity becomes 3.20 m/s. What is the change in kinetic energy of the cart? How much work was done on the cart?

  8. Example Problem Ki = ½(15.0 kg)(7.50 m/s)2 = 422 J Kf = ½(15.0kg)(3.20m/s)2 = 76.8 J DK =Kf - Ki = 76.8 J – 422 J = -345.2 J W=DK = - 345.2J

  9. Example Problem Eesha weighs 505 N. She walks down a flight of stairs at Solon High School to a level 5.50 m below her starting point. What is the change in Eesha’s potential energy?

  10. Example Problem PE = (mg)h = (Fg)h = (505N)(-5.50m) = -2780 J DPE = PEf - PEi = -2780 J – 0 = -2780 J

  11. Example Problem • K = 1/2mv2 • K α m • Double mass  K doubles • Cut mass in ½  K is cut in 1/2 • K α v2 • double v  (2v)2 = 4v  K quadruples • cut v ½  (1/2v)2 = 1/4v  K is cut by 1/4

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