Does The Air Resistance On The Coffee Filter Obey A Linear Or Quadratic Relationship For Velocity?

22.5 Conservation of mechanical energy (ESAHO)

Conservation of Energy: The Law of Conservation of Energy: Energy cannot be created or destroyed, just is merely inverse from one class into another.

So far we have looked at ii types of energy: gravitational potential free energy and kinetic energy. The sum of the gravitational potential free energy and kinetic energy is chosen the mechanical energy. In a closed system, one where there are no external dissipative forces acting, the mechanical free energy volition remain constant. In other words, it volition non alter (get more than or less). This is called the Law of Conservation of Mechanical Energy.

In problems involving the use of conservation of energy, the path taken by the object tin exist ignored. The only of import quantities are the object's velocity (which gives its kinetic free energy) and height to a higher place the reference point (which gives its gravitational potential free energy).

Conservation of mechanical energy: Police force of Conservation of Mechanical Energy: The full corporeality of mechanical energy, in a airtight organization in the absence of dissipative forces (e.g. friction, air resistance), remains constant.

This means that potential energy tin can become kinetic energy, or vice versa, but energy cannot "disappear". For example, in the absence of air resistance, the mechanical energy of an object moving through the air in the Earth'due south gravitational field, remains constant (is conserved).

Using the law of conservation of energy (ESAHP)

Mechanical energy is conserved (in the absenteeism of friction). Therefore nosotros can say that the sum of the ${E}_{P}$ and the ${Eastward}_{Yard}$ anywhere during the motility must exist equal to the sum of the the ${Due east}_{P}$ and the ${East}_{K}$ anywhere else in the motion.

We can at present employ this to the example of the suitcase on the closet. Consider the mechanical energy of the suitcase at the top and at the bottom. We tin can say:

\begin{align*} {E}_{M1} & = {East}_{M2} \\ {E}_{P1} + {East}_{K1} & = {East}_{P2} + {E}_{K2} \\ mgh + \frac{1}{ii}thou{5}^{ii} & = mgh + \frac{one}{2}thousand{5}^{2} \\ \left(\text{ane}\text{ kg}\right)\left(\text{9,8}\text{ m·southward$^{-2}$}\correct)\left(\text{two}\text{ one thousand}\correct) + 0 & = 0 + \frac{one}{2}\left(\text{ane}\text{ kg}\correct)\left({5}^{2}\right) \\ \text{xix,half dozen} & = \frac{one}{2}\left({five}^{2}\correct) \\ {v}^{2} & = \text{39,2}\text{ thousand$^{2}$·s$^{-two}$} \\ v & = \text{vi,26}\text{ m·due south$^{-one}$} \terminate{marshal*}

The suitcase will strike the ground with a velocity of $\text{vi,26}$ $\text{m·southward$^{-1}$}$.

From this we see that when an object is lifted, similar the suitcase in our example, it gains potential free energy. As information technology falls back to the footing, it will lose this potential energy, but proceeds kinetic energy. We know that energy cannot be created or destroyed, but only changed from one form into another. In our example, the potential energy that the suitcase loses is changed to kinetic free energy.

The suitcase will have maximum potential energy at the top of the cupboard and maximum kinetic energy at the bottom of the closet. Halfway downwards it volition have half kinetic energy and half potential energy. As it moves down, the potential energy will be converted (changed) into kinetic energy until all the potential free energy is gone and only kinetic energy is left. The $\text{19,6}$ $\text{J}$ of potential energy at the acme will get $\text{xix,6}$ $\text{J}$ of kinetic free energy at the lesser.

Conversion of energy

Materials

A length of plastic pipe with diameter approximately 20 mm, a marble, some masking tape and a measuring tape.

To do (1)

First put i finish of the pipage on the tabular array top so that it is parallel to the top of the table and tape it in position with the masking tape.

Lift the other terminate of the pipe upwards and hold it at a steady meridian not too high to a higher place the table.

Measure the vertical height from the table top to the meridian opening of the piping.

Now put the marble at the top of the pipe and let information technology go then that it travels through the piping and out the other stop.

Questions

What is the velocity (i.east. fast, slow, not moving) of the marble when you first put it into the acme of the pipe and what does this hateful for its gravitational potential and kinetic free energy?
What is the velocity (i.eastward. fast, slow, not moving) of the marble when it reaches the other end of the pipe and rolls onto the desk-bound? What does this mean for its gravitational potential and kinetic free energy?

To practise (2)

Now elevator the pinnacle of the piping equally high equally it volition go.

Measure the vertical height of the peak of the pipe above the table top.

Put the marble into the elevation opening and let it gyre through the pipe onto the table.

Questions

What is the velocity (i.due east. fast, slow, not moving) of the marble when you put information technology into the superlative of the pipe, and what does this mean for its gravitational potential and kinetic energy?
Compared to the starting time attempt, what was unlike about the height of the meridian of the tube? How practise yous recall this affects the gravitational potential energy of the marble?
Compared to your beginning attempt, was the marble moving faster or slower when information technology came out of the bottom of the piping the 2nd time? What does this mean for the kinetic energy of the marble?

The activity with the marble rolling down the pipe shows very nicely the conversion betwixt gravitational potential energy and kinetic energy. In the start instance, the pipe was held relatively low and therefore the gravitational potential free energy was too relatively low. The kinetic energy at this point was zero since the marble wasn't moving yet. When the marble rolled out of the other end of the pipe, it was moving relatively slowly, and therefore its kinetic energy was also relatively low. At this bespeak its gravitational potential energy was zip since it was at zero height above the table superlative.

In the second example, the marble started off above and therefore its gravitational potential free energy was higher. Past the time information technology got to the bottom of the piping, its gravitational potential energy was nothing (goose egg pinnacle to a higher place the tabular array) but its kinetic energy was loftier since information technology was moving much faster than the first time. Therefore, the gravitational potential energy was converted completely to kinetic energy (if nosotros ignore friction with the piping).

In the case of the pipe being held college, the gravitational potential free energy at the start was higher, and the kinetic energy (and velocity) of the marble was higher at the end. In other words, the full mechanical energy was college and and just depended on the elevation yous held the pipe above the table summit and not on the altitude the marble had to travel through the pipe.

Worked example vii: Using the Law of Conservation of Mechanical Energy

During a flood a tree trunk of mass $\text{100}$ $\text{kg}$ falls downward a waterfall. The waterfall is $\text{five}$ $\text{m}$ high.

If air resistance is ignored, summate:

the potential energy of the tree trunk at the top of the waterfall.
the kinetic energy of the tree trunk at the bottom of the waterfall.
the magnitude of the velocity of the tree trunk at the bottom of the waterfall.

Analyse the question to determine what information is provided

The mass of the tree trunk $m = \text{100}\text{ kg}$
The top of the waterfall $h = \text{5}\text{ chiliad}$.

These are all in SI units so nosotros do not have to convert.

Analyse the question to determine what is existence asked

Potential energy at the peak
Kinetic energy at the bottom
Velocity at the bottom

Calculate the potential energy at the peak of the waterfall.

\begin{align*} {East}_{P} & = mgh \\ & = \left(\text{100}\text{ kg}\correct)\left(\text{9,8}\text{ m·s$^{-2}$}\right)\left(\text{5}\text{ 1000}\right) \\ & = \text{4 900}\text{ J} \end{align*}

Calculate the kinetic energy at the bottom of the waterfall.

The full mechanical energy must be conserved.

\[{E}_{K1} + {East}_{P1} = {E}_{K2} + {E}_{P2}\]

Since the trunk'south velocity is zero at the acme of the waterfall, ${E}_{K1}=0$.

At the bottom of the waterfall, $h = \text{0}\text{ one thousand}$, then ${Eastward}_{P2}=0$.

Therefore ${Due east}_{P1} = {E}_{K2}$ or in words:

The kinetic energy of the tree torso at the bottom of the waterfall is equal to the potential energy it had at the elevation of the waterfall. Therefore ${E}_{K} = \text{iv 900}\text{ J}$

Calculate the velocity at the bottom of the waterfall.

To calculate the velocity of the tree trunk we need to employ the equation for kinetic free energy.

\begin{align*} {E}_{1000} & = \frac{ane}{2}m{v}^{2} \\ \text{four 900} & = \frac{i}{two}\left(\text{100}\text{ kg}\right)\left({5}^{2}\right) \\ 98 & = {v}^{ii} \\ v & = \text{nine,899...}\text{ m·s$^{-1}$} \\ v & = \text{ix,90}\text{ m·s$^{-1}$} \end{align*}

Worked instance 8: Pendulum

A $\text{2}$ $\text{kg}$ metal brawl is suspended from a rope as a pendulum. If information technology is released from betoken A and swings downward to the bespeak B (the lesser of its arc):

bear witness that the velocity of the ball is independent of its mass,
calculate the velocity of the ball at bespeak B.

Analyse the question to make up one's mind what information is provided

The mass of the metal ball is $m = \text{2}\text{ kg}$
The change in height going from point A to point B is $h = \text{0,5}\text{ thousand}$
The ball is released from betoken A so the velocity at signal, ${v}_{A} = \text{0}\text{ m·s$^{-1}$}$.

All quantities are in SI units.

Analyse the question to determine what is being asked

Prove that the velocity is independent of mass.
Find the velocity of the metal brawl at signal B.

Apply the Law of Conservation of Mechanical Energy to the situation

Since in that location is no friction, mechanical energy is conserved. Therefore:

\begin{align*} {E}_{M1} & = {E}_{M2} \\ {Due east}_{P1} + {Eastward}_{K1} & = {E}_{P2} + {Eastward}_{K2} \\ mg{h}_{i} + \frac{one}{2}m{\left({v}_{1}\right)}^{2} & = mg{h}_{ii} + \frac{1}{2}m{\left({v}_{2}\correct)}^{2} \\ mg{h}_{1} + 0 & = 0 + \frac{1}{ii}m{\left({v}_{ii}\right)}^{2} \\ mg{h}_{1} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} \finish{align*}

The mass of the ball $m$ appears on both sides of the equation so it can be eliminated so that the equation becomes:

\begin{align*} g{h}_{one} & = \frac{1}{two}{\left({v}_{two}\right)}^{2} \\ 2g{h}_{1} & = {\left({v}_{two}\correct)}^{2} \terminate{align*}

This proves that the velocity of the ball is independent of its mass. It does not affair what its mass is, information technology will always have the same velocity when it falls through this height.

Calculate the velocity of the ball at point B

Nosotros can use the equation above, or exercise the calculation from "kickoff principles":

\begin{marshal*} {\left({v}_{two}\right)}^{ii} & = 2g{h}_{ane} \\ {\left({v}_{2}\right)}^{2} & = \left(ii\correct)\left(\text{nine,8}\text{ thousand·s$^{-2}$}\right)\left(\text{0,5}\text{ m}\right) \\ {\left({v}_{2}\correct)}^{two} & = \text{ix,8}\text{ m$^{2}$·s$^{-2}$} \\ {5}_{two} & = \sqrt{\text{9,8}\text{ m$^{2}$·s$^{-ii}$}} \\ {five}_{ii} & = \text{3,13}\text{ one thousand·s$^{-one}$} \end{align*}

Alternatively you can do:

\begin{align*} {E}_{K1} + {Eastward}_{P1} & = {Due east}_{K2} + {Eastward}_{P2} \\ mg{h}_{1} + \frac{1}{2}m{\left({5}_{1}\correct)}^{ii} & = mg{h}_{two} + \frac{one}{2}m{\left({v}_{ii}\right)}^{2} \\ mg{h}_{1} + 0 & = 0 + \frac{one}{2}1000{\left({v}_{two}\correct)}^{2} \\ {\left({v}_{2}\right)}^{ii} & = \frac{2mg{h}_{1}}{m} \\ {\left({v}_{ii}\correct)}^{two} & = \frac{2\left(\text{2}\text{ kg}\correct)\left(\text{9,8}\text{ yard·southward$^{-2}$}\correct)\left(\text{0,5}\text{ m}\right)}{\text{ii}\text{ kg}} \\ {v}_{2} & = \sqrt{\text{9,8}\text{ m$^{two}$·s$^{-ii}$}} \\ {v}_{ii} & = \text{3,xiii}\text{ m·s$^{-ane}$} \end{align*}

Worked example 9: The roller coaster

A roller coaster ride at an amusement park starts from residual at a height of $\text{50}$ $\text{thou}$ above the ground and quickly drops down forth its track. At some betoken, the track does a full 360° loop which has a summit of $\text{xx}$ $\text{k}$, before finishing off at ground level. The roller coaster railroad train itself with a total load of people on information technology has a mass of $\text{850}$ $\text{kg}$.

Roller coaster

If the roller coaster and its track are frictionless, summate:

the velocity of the roller coaster when it reaches the top of the loop
the velocity of the roller coaster at the bottom of the loop (i.e. ground level)

Analyse the question to decide what information is provided

The mass of the roller coaster is $m = \text{850}\text{ kg}$
The initial superlative of the roller coaster at its starting position is ${h}_{i} = \text{50}\text{ m}$
The roller coaster starts from residue, so its initial velocity ${five}_{one} = \text{0}\text{ m·s$^{-1}$}$
The height of the loop is ${h}_{ii} = \text{twenty}\text{ m}$
The height at the lesser of the loop is at ground level, ${h}_{3} = \text{0}\text{ m}$

We do not need to convert units equally they are in the right form already.

Analyse the question to determine what is beingness asked

the velocity of the roller coaster at the peak of the loop
the velocity of the roller coaster at the bottom of the loop

Calculate the velocity at the tiptop of the loop

From the conservation of mechanical energy, Nosotros know that at whatever two points in the organisation, the full mechanical energy must be the same. Permit'due south compare the state of affairs at the start of the roller coaster to the situation at the meridian of the loop:

\brainstorm{align*} {E}_{M1} & = {E}_{M2} \\ {Eastward}_{K1} + {East}_{P1} & = {East}_{K2} + {E}_{P2} \\ 0 + mg{h}_{ane} & = \frac{one}{2}m{\left({5}_{2}\correct)}^{2} + mg{h}_{2} \end{align*}

We can eliminate the mass, $thousand$, from the equation by dividing both sides by $m$.

\brainstorm{align*} thousand{h}_{1} & = \frac{1}{2}{\left({v}_{2}\right)}^{two} + grand{h}_{2} \\ {\left({v}_{2}\right)}^{2} & = two\left(g{h}_{1} - g{h}_{2}\right) \\ {\left({v}_{2}\correct)}^{2} & = 2\left(\left(\text{9,8}\text{ one thousand·due south$^{-2}$}\right)\left(\text{50}\text{ m}\right)-\left(\text{9,8}\text{ k·due south$^{-2}$}\right)\left(\text{20}\text{ m}\right)\right) \\ {five}_{two} & = \text{24,25}\text{ thousand·due south$^{-1}$} \end{align*}

Calculate the velocity at the bottom of the loop

Again we can use the conservation of energy and the full mechanical energy at the bottom of the loop should be the same as the full mechanical energy of the system at whatever other position. Let's compare the situations at the commencement of the roller coaster'due south trip and the bottom of the loop:

\brainstorm{marshal*} {E}_{M1} & = {Eastward}_{M3} \\ {Due east}_{K1} + {E}_{P1} & = {Due east}_{K3} + {E}_{P3} \\ \frac{ane}{2}{chiliad}_{1}{\left(0\right)}^{2} + mg{h}_{ane} & = \frac{1}{2}1000{\left({v}_{iii}\right)}^{two} + mg\left(0\right) \\ mg{h}_{i} & = \frac{1}{two}m{\left({v}_{3}\correct)}^{two} \\ {\left({v}_{iii}\right)}^{ii} & = 2g{h}_{ane} \\ {\left({v}_{3}\correct)}^{2} & = 2\left(\text{9,eight}\text{ one thousand·south$^{-ii}$}\right)\left(\text{l}\text{ thou}\right) \\ {five}_{3} & = \text{31,30}\text{ 1000·due south$^{-1}$} \end{marshal*}

Worked example 10: An inclined plane

A mount climber who is climbing a mount in the Drakensberg during winter, by mistake drops her water canteen which then slides $\text{100}$ $\text{g}$ down the side of a steep icy gradient to a point which is ten m lower than the climber's position. The mass of the climber is $\text{sixty}$ $\text{kg}$ and her h2o bottle has a mass of $\text{500}$ $\text{g}$ .

If the canteen starts from rest, how fast is it travelling past the time it reaches the lesser of the slope? (Neglect friction.)
What is the total alter in the climber'south potential energy as she climbs downwards the mount to fetch her fallen water canteen? i.e. what is the deviation betwixt her potential energy at the top of the gradient and the bottom of the slope?

Analyse the question to determine what data is provided

the distance travelled by the water bottle downward the slope, $d = \text{100}\text{ thousand}$
the difference in height betwixt the starting position and the final position of the water bottle is $h = \text{10}\text{ m}$
the bottle starts sliding from residual, so its initial velocity is ${v}_{i} = \text{nine,8}\text{ thousand·s$^{-ane}$}$
the mass of the climber is $\text{sixty}$ $\text{kg}$
the mass of the water bottle is $\text{500}$ $\text{one thousand}$. We need to convert this mass into $\text{kg}$: $\text{500}\text{ thousand} = \text{0,five}\text{ kg}$

Analyse the question to determine what is being asked

What is the velocity of the h2o bottle at the bottom of the gradient?
What is the difference between the climber's potential energy when she is at the top of the gradient compared to when she reaches the bottom?

Calculate the velocity of the water bottle when it reaches the bottom of the slope

\begin{align*} {E}_{M1} & = {E}_{M2} \\ {E}_{K1} + {E}_{P1} & = {E}_{K2} + {E}_{P2} \\ \frac{ane}{2}1000{\left({v}_{1}\right)}^{2} + mg{h}_{ane} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} + mg{h}_{2} \\ 0 + mg{h}_{i} & = \frac{1}{two}yard{\left({v}_{2}\correct)}^{2} + 0 \\ {\left({v}_{2}\right)}^{two} & = \frac{2mgh}{thousand} \\ {\left({5}_{two}\right)}^{2} & = 2gh \\ {\left({v}_{ii}\correct)}^{2} & = \left(2\right)\left(\text{9,8}\text{ thousand·south$^{-2}$}\right)\left(\text{x}\text{ 1000}\right) \\ {5}_{2} & = \text{xiv}\text{ m·south$^{-1}$} \finish{marshal*}

Annotation: the distance that the bottle travelled (i.eastward. $\text{100}$ $\text{m}$) does not play any part in calculating the energies. It is only the height difference that is important in calculating potential free energy.

Summate the difference between the climber's potential free energy at the tiptop of the slope and her potential energy at the lesser of the slope

At the elevation of the slope, her potential free energy is:

\begin{marshal*} {E}_{P1} & = mg{h}_{1} \\ & = \left(\text{60}\text{ kg}\right)\left(\text{nine,8}\text{ m·s$^{-ii}$}\right)\left(\text{ten}\text{ m}\right) \\ & = \text{5 800}\text{ J} \end{align*}

At the bottom of the gradient, her potential free energy is:

\begin{align*} {Due east}_{P1} & = mg{h}_{i} \\ & = \left(\text{sixty}\text{ kg}\correct)\left(\text{ix,eight}\text{ m·s$^{-2}$}\right)\left(\text{0}\text{ m}\right) \\ & = \text{0}\text{ J} \end{marshal*}

Therefore the difference in her potential energy when moving from the top of the slope to the bottom is:

\[{Eastward}_{P1} - {East}_{P2} = \text{5 880} - 0 = \text{5 880}\text{ J}\]

Potential energy

Textbook Exercise 22.3

A tennis ball, of mass $\text{120}$ $\text{kg}$ , is dropped from a height of $\text{5}$ $\text{m}$ . Ignore air friction.

What is the potential energy of the brawl when information technology has fallen $\text{iii}$ $\text{m}$ ?
What is the velocity of the ball when information technology hits the ground?

Solution not withal bachelor

A ball rolls downward a hill which has a vertical height of $\text{15}$ $\text{k}$ . Ignoring friction, what would be the

gravitational potential free energy of the ball when it is at the top of the hill?
velocity of the brawl when it reaches the bottom of the hill?

Solution not yet available

A bullet, mass $\text{50}$ $\text{g}$ , is shot vertically up in the air with a muzzle velocity of $\text{200}$ $\text{one thousand·s$^{-1}$}$. Utilize the Principle of Conservation of Mechanical Free energy to decide the height that the bullet will reach. Ignore air friction.

Solution not yet available

A skier, mass $\text{l}$ $\text{kg}$, is at the top of a $\text{6,4}$ $\text{m}$ ski slope.

Determine the maximum velocity that she can attain when she skis to the lesser of the slope.
Practice you lot think that she will reach this velocity? Why/Why not?

Solution not yet available

A pendulum bob of mass $\text{i,5}$ $\text{kg}$, swings from a top A to the lesser of its arc at B. The velocity of the bob at B is $\text{4}$ $\text{m·southward$^{-ane}$}$. Calculate the height A from which the bob was released. Ignore the effects of air friction.

Solution not nevertheless available

Prove that the velocity of an object, in costless fall, in a airtight system, is contained of its mass.

Solution not yet available