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Section 5–2: Energy. Physics Coach Kelsoe Pages 164 – 172. Objectives. Identify several forms of energy. Calculate kinetic energy for an object. Apply the work-kinetic energy theorem to solve problems. Distinguish between kinetic and potential energy.

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### Section 5–2: Energy

Physics

Coach Kelsoe

Pages 164–172

• Identify several forms of energy.

• Calculate kinetic energy for an object.

• Apply the work-kinetic energy theorem to solve problems.

• Distinguish between kinetic and potential energy.

• Classify different types of potential energy.

• Calculate the potential energy associated with an object’s position.

• Remember from Section 1 that Wnet=Fnetd, which can also be said as Wnet=maΔx.

• From our one-dimensional motion section, we learned that:

• vf2 = vi2 + 2aΔx, so then

• aΔx = vf2 - vi2/2

• Substituting this result into the Work formula gives us

• Wnet = m(vf2 - vi2/2) OR

• Wnet = ½mvf2 - ½mvi2

• The quantity of ½mv2 in physics is called kinetic energy.

• Kinetic energy is the energy of an object due to it’s motion.

• KE = ½mv2

• KE is a scalar quantity and is measured in joules. It is dependent upon two quantities: mass and speed.

• Consider which has more KE, a bowling ball rolling down a hill or a volleyball rolling down the hill at the same speed.

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

• 1. Identify givens and unknowns:

• mb = 7.00 kg

• mp = 0.00245 kg

• vb = 3.00 m/s

• vp = ?

• 2. Choose the correct equation

• KE = ½mv2

• Knowing that the KEs must equal, then, KEb = KEp

• ½mbvb2= ½mpvp2

• 3. Substitute values into equation.

• ½(7.00 kg)(3.00 m/s)2= ½(.00245 kg)vp2

• vp2 = √ ½(7.00 kg)(3.00 m/s)2/ ½(.00245 kg)

• vp = 160 m/s

• I would definitely say this is a wee bit fast for a ping pong ball!

• The work-kinetic energy theorem states that net work done on a body equals its change in kinetic energy.

• Wnet = ΔKE –OR– Wnet = ½mvf2 - ½mvi2

• When this theorem is used, all the forces that do work on an object must be included.

• On a frozen pond, a person kicks a 10.0 kg 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?

• 1. Define givens and unknowns:

• m = 10.0 kg

• vi = 2.2 m/s

• vf = 0.0 m/s

• μk = 0.10

• d = ?

• Choose an equation or situation: This problem can be solved using the definition of work and the work-kinetic energy theorem.

• Wnet = Fnetd(cos θ)

• The net work done on the sled is provided by the force of kinetic friction.

• Wnet = Fkd(cos θ)

• Wnet = μkFnd(cos θ)

• We also know that Wnet = KEf – KEi and the final KE is zero, so in effect,

• Wnet = - KEi

• Wnet = - ½mvi2

• μkFnd(cos θ) = - ½mvi2

• 3. Solve

• Manipulating the formula above to solve d.

• d = -vi2/μkg(cos θ)

• d = -(2.2m/s)2/(0.10)(9.81m/s2)(cos 180º)

• d = 2.5 m

• *NOTE* We used 180º because the COF was in the opposite direction of the force applied.

• Objects in motion have kinetic energy, but it can also have other forms of energy.

• Potential energy is the energy associated with an object because of the position, shape, or condition of the object.

• Potential energy depends not only on the properties of an object, but also on the object’s interaction with its environment.

• Gravitational potential energy (PEg) is the potential energy stored in the gravitational fields of interacting bodies.

• Gravitational potential energy depends on height from a zero level.

• PEg = mgh

• Gravitational PE = mass x free-fall acceleration x height

• An object does not lose its PEg even as it falls.

• Imagine an ink pen falling off your desk. Although as it falls it has kinetic energy (KE), it still has some gravitational potential energy (PEg) until it hits the floor.

• For an object falling from rest, its KE will never be greater than its PEg.

• ΣEnergy = PEg + KE

• Elastic potential energy (PEelastic) is the energy available for use when a deformed elastic object returns to its original configuration.

• PEelastic = ½kx2

• PEelastic = ½(spring constant)(distance compressed or stretched)2

• Elastic potential energy is stored in any compressed or stretched object, such as a spring or the stretched strings of a tennis racket or guitar.

• The symbol k is called the spring constant, or force constant. It is measured in units of N/m.

• The length of a spring when no external forces are acting on it is called the relaxed length of the spring.

• When an external force compresses or stretches the spring, elastic potential energy is stored in the spring.

• A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. 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 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?

• 1. Identify givens and unknowns:

• 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

• PEtotal = ?

• The zero level for gravitational potential energy is chosen to be at the surface of the water. The total potential energy is the sum of the gravitational and elastic potential energy.

• PEtotal = PEg + PEelastic

• PEg = mgh

• PEelastic = ½kx2

• 3. Calculate

• PEg = (70.0 kg)(9.81 m/s2)(6.0 m)

• PEg = 4100 J

• PEelastic = ½(71.8 N/m)(29.0 m)2

• PEelastic = 30,200 J

• PEtotal = 4100 J + 30,200 J

• PEtotal = 34,300 J

• Kinetic energy

• Work-kinetic energy theorem

• Potential energy

• Gravitational potential energy

• Elastic potential energy

• Spring constant