Equations for mechanical potential and kinetic energy

By Joseph Moore,2014-04-17 22:59
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Equations for mechanical potential and kinetic energy

Equations for mechanical potential and kinetic energy

    1. Imagine pushing on a moveable object on a frictionless surface with a device that converts gasoline’s potential energy to kinetic energy of motion of the object.


    Force ?? _________________


    2. The amount of gasoline needed (potential energy used) depends upon how hard the push is and the distance being pushed. Think of yourself pushing….how tired you feel refers to how much PE you have lost. A small force for a long distance would produce the same feeling as a big force moving a small distance. The shape, color, mass, speed etc of the object only affect the interaction through the force applied or distance moved. To create an equation for Kinetic energy gained we can use Newton’s


     F d = loss PE = gain KE = ma d using F=ma

     2 22222Fd= mad = m ( VV ) d using kinematics equation V = V + 2ad or (V-V)/2a = d 212121

    2222So that Fd = mad = m a (V V )/ 2a or ? mV - ? mV = KE= KE KE 212121

    2 PE ==================== KE and therefore KE must be ? mV for a moving object!

    3. IF the force applied is NOT horizontal, if the surface is NOT frictionless the equation becomes:

    F ; d = KE + heat energy or /PE/ = KE + heat energy

     Dot product Absolute value

If PE will be negative if PE is lost and KE and heat will be positive if the object speeds up.

Mathematically the equation now is: - PE = KE + Heat

Or in a more useful form: { 0 = PE + KE + Heat }

     2Most texts use: F d = ? mV which is very confusing conceptually if easier mathematically. Even F d = KE which is better obscures the role of conservation of energy. For a closed system F ; d = KE (of object pushed) + PE (of object pushed) + Heat ( Ff x d)

     PE being lost by pusher which is (-) is changed to (+) to equal the sum of all the terms to the right.

     For example: A pusher does 20 J of work on a block sliding up a hill with friction: The results may be: 20J = 12 J KE increase + 4 J Pe grav rising + 6 J heat from friction or

     0 = -20 J + 20J which expresses the whole transaction in terms

     of conservation of energy

     Gravitational potential energy changes

    4. When lifting a mass a vertical distance h the typical equation given is: PE = mgh

     Where mg represents the force applied upward and h the distance.

Several conceptual problems arise.

     st1 F = mg represents the “average upward force” to take a mass from vertical rest to a higher rest

    position. The upward force is greater than mg accelerating the mass upward getting it moving but its

    less than mg slowing it down to rest.

     nd2 Since gravity varies as the inverse square of the distance to earth center the value of g actually gets

    very slightly smaller with height.

     rd3 PE = mgh implies if the object is not lifted and h is zero the object has no PE which is false and conceptually confusing. All objects have PE and it just changes when they are lifted.

     th4 One can calculate the gravitational potential energy of a mass but only by using the equation:

     PE = G Mass x M earth / Re at the surface. Mgh should be written mgh or (mg{h h}) 21

     where h stands for the radius of the earth Re. h can be replaced with d. 1

     A reasonable conceptual compromise can be PE grav = mgh or mgd

When an object moves further from the earth’s center PE rises and PE is (+) and

    when it moves closer to earth PE is (-)

     th5 How much work is done lifting an object down instead of up?

     If the mass starts and ends with zero velocity then it takes just as much work to lift it down as to lift it up! If the system includes the lifter and the mass the person is always giving off energy in the form of heat to the air. Suppose you take 10 s to lift something up and then 10 s to lift it down: Example:

    Give a mass 10 J PE by lifting and at the same time give 3 J of heat to the air. Your body has lost 13 J of energy. When you lift it down the mass loses 10 J of PE , your body loses 3 J and l0 J of heat. If we ignore the 3 J of body heat each way the equation becomes:

     F d = ΔPE for the block system alone 1

     Work done by body on system 10 J = -10 J energy is not conserved?

     Some texts suggest -Fd when lowering to make math correct; that is a +Fd when putting energy into a system and a F d when taking energy out. This works ok but doesn’t clarify “how tired we get “ …..our change in PE when we do it.

     The following shows how to keep track of system including air, pusher and block

     conservation of energy theme …lifting then lowering the block system

     0 = ΔPE + ΔKE+ ΔPE blockblockbody

     Lifting the block 0 = + 10 J + 0 - 10 seems ok?


     Including air 0= +10 J + 0 J 3 J -10J + + 3 J

     Body loses 13 J energy 0= ΔPE + ΔKE+ ΔPE + KE blockblockbodyair

     Lowering the block 0 = ΔPE + ΔKE+ ΔPE + KE blockblockbodyair

     0= -10 J + 0 - 10 J + 20 J

     including air 0 = -10 J + 0 + -3 J -10 J + 23 J

     Body loses 13 J energy

     Definition of Work

    1. Most texts define work as simply: W= Force * distance This works ok to

    describe energy transferred into a system capable of movement. But students equate work with “tired” and so pushing on a wall makes them tired but does no “work”. This is confusing

2. A better definition of work in encludes the words “to” and “on”.

    Work is the transfer of energy from one system “ to” another and the

    work is done “on” a particular thing. When you are pushing on the wall

    you are doing work “on” the air because you are transferring energy “to” it.

    Work done “on” a mass moving it doesn’t have to include the air thus F d

    for work done on the mass. The work done

    on the mass alone may manifest itself in

     ΔPE, ΔKE, Δ KE , ΔPE , ΔPE………. gravitational mechanical heatelectricaletc

     For a block being pushed across a rough surface that slopes upward:

    F * d = Δ PE + Δ KE + Δ KE gravitational mechanical heat

     goes up goes up goes up

    22= mgΔh + { ? MV ? MV} + F d + etc 21friction F * d

work done on system of the block = energy absorbed in several forms by the block system

Frictional heat can be represented by { F * d = F d } by analysis of pushing an object at constant speed on appliedfriction

    a level surface.

    This form of the conservation of energy relates well to pushing on a moveable system and for most problems encountered in the text this calculates what happens to the block.

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