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Physics 170 - MechanicsLecture 17

Potential Energy

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Conservative Force: Gravity

Work done by gravity on a closed path is zero:

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Nonconservative Force: FrictionWork done by friction on a closed path is not zero:

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Conservative Forces and Paths The work done by a conservative force is zero on any closed path:

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Conservative andNonconservative Forces

Conservative force: the work it does is stored in the form of energy that can be released at a later time

Examples of a conservative force: gravity. spring

Example of a nonconservative force: friction

Also: the work done by a conservative force in moving an object around a closed path is zero;this is not true for a nonconservative force.

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Heat: Calories vs. joules Heat energy is often produced by nonconservative forces. Energy, particularly heat energy, is sometimes quantified in units of calories, defined as the heat energy required to raise 1 g (= 1 ml = 1 cm3) of water by 1 degree Celsius. 1 cal = 4.186 J.

The equivalent SI unit is the kilocalorie, denoted by Cal or kcal, which is the heat energy required to raise 1 kg (or 1 liter) of water by 1 degree Celsius. 1 Cal = 1 kcal = 4.186 x 103 J. This is the unit often used to indicate the energy content of food, and is measured by burning the food in a calorimeter and measuring the heat output.

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Conservative Forces & Potential Energy

Potential energy is an energy of position. The system has a unique value of potential energy when the object is at A, a different value when the object is at B, etc. Thus, the net change in potential energy is ΔU=UB-UA, and is the same whether the object moves form A to B along Path 1 or along Path 2.

Potential energy is transformed to kinetic energy with ΔK=-ΔU. ΔU is independent of path, and ΔK is independent of path.

The change in the particle’s kinetic energy is related to the amount of work done on the particle by force F. According to the work-kinetic-energy theorem, ΔK=W. Because ΔK is independent of path, the work W done by the force in moving from A to B must also be independent of path.

A potential energy can be associated with any conservative force.

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Work Done byConservative Forces: GravityGravitational potential energy:

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Work Done byConservative Forces: Gravity

If we pick up a ball and put it on the shelf, we have done work on the ball. We can get that energy back if the ball falls back off the shelf; in the meantime, we say the energy is stored as potential energy.

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Example: A Falling Bottle A 0.350 kg bottle falls from rest from a shelf that is 1.75 m above the floor.

(a) Find the potential energy of the bottle when the bottle is on the shelf.

(b) Find the kinetic energy of the bottle just before impact with the floor.

= (yf =0)

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Example: Pike’s Peak or Bust An 82.0 kg mountain climber is in the final stage of the ascent of Pike’s Peak, which 4,301 m above sea level.

(a) What is the change in gravitational potential energy as the climber gains the last 100.0 m of altitude? Use U=0 at sea level.

(b) Do the same calculation with U=0 at the top of the peak.

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Example: A Mountain Bar A candy bar called the Mountain Bar has an energy content when metabolized of 212 Cal = 212 kcal. This is equivalent to 8.87 x 105 J.

If an 81.0 kg mountain climber eats a Mountain Bar and magically converts all of its energy content into gravitational potential energy, how much altitude Δy should he be able to gain?

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Conservative Forcesand Potential Energy

A potential energy can be associated with any conservative force.

Both are location-dependent and reversible potential energies.

Note that friction is not a conservative force and is irreversible.

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Example:The Speed of a Falling Rock

A rock of 1.0 Kg is released from rest. Use both Betty’s and Bill’s perspectives to calculate its speed just before it hits the ground.

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Example: The Speed of a Sled Christine runs forward with her sled at 2.0 m/s. She hops onto the sled at the top of a 5.0 m high, very slippery slope. What is her speed at the bottom?

½mv12+mgy1 = ½mv0

2+mgy0v1 = [v0

2 + 2gy1]½

= [(2.0 m/s)2+2(9.80 m/s2)(5.0 m)]½ = 10.1 m/s

K1 + Ug1 = K0 + Ug0

Notice that the steepness of the slope and/or whether it has bumps and dips does not matter in determining the answer. Only the change in height and the initial speed are relevant to the answer.