Water Slide Collisions!

Water Slide Collisions

Nitro Racer – Water Country USA Williamsburg, Virginia

http://www.watercountryusa.com/wc/Explore/Rides.aspx?id=1098

http://pennrecord.com/news/collision-on-nitro-racer-basis-for-injury-claim-against-seaworld/

http://www.youtube.com/watch?v=YmZbJg4pi7o

Vertical Drop:  150ft (45.7m)

Length of Slide 320ft (97.5m)

Mass of Kid:  40kg

Mass of Father:  80kg

Collision occurs over .05s

Potential Energy = mgh

Kinetic Energy = 1/2mv²

Kid drops to vertical elevation of 25ft (7.6m)

Kid PE = (40kg)(9.8m/s²)(7.6) = 2980J

Man Drops to vertical elevation of 25ft (7.6m) before collision

Man PE at 95.5m = (80kg)(9.8m/s²)(45.7) = 35800J

Man PE at 7.6m = (80kg)(9.8m/s²)(7.6) = 5960J

Man KE at 7.6m = -mgh + mgh - mmg * sin(q)(d) =  Velocity of man before collision = -(80kg * 9.8m/s ² * 7.6m) + (80kg * 9.8m/s² * 45.7m) – (.2 *80kg*  9.8m/s²)(sin 28)(97.5m-16.2m) = 24.4m/s

The Collision!

M₁V₁ + M₂V₂ = v(M₁+M₂)

(80kg)(24.4m/s) + (40kg)(0m/s) = (120kg)(v) v= 16.2 m/s

Impulse = m∆v

Impulse = 120kg(16.2m/s-24.4m/s) = -984kg*m/s

Force = Impulse / ∆t

Force = 984kg*m/s / .05s

19,700N Very possible that serious injuries were sustained

30,000N needed to break some bones such as the patella

Chances are that the bones would not break!

The Force of the Frankenstorm

We all know Hurricane Sandy rocked the East Coast over the past couple of days, and I was curious to check out some of the statistics for the Superstorm. I found that, along with huge amounts of rain and snowfall, gusts of wind reached as high as 94 mph (42 m/s) in places like Eaton's Neck, NY.

http://www.scientificamerican.com/article.cfm?id=the-stats-are-in-superstorm-sandy

This reminded me of a homework problem we'd done a little while back, with wind being strong enough to blow over a person. So, I wanted to look at situation that could definitely occur in a hurricane: wind blowing over an entire car.
I assumed that the wind is approaching the car from the side, so that the car is basically a rectangle. I found that the average length of a sedan is about 4.6 m, and the height is 1.5 m, and the mass is 1,500 kg. I also assumed the wind is striking the car at a rate of 66 kg/s per square meter and comes to rest. I wanted to: A. Find the approximate force of the wind on the car, and B.Compare this force to the typical maximum friction force between a car and the wet road.

A)
1. First find the area of the car on which the wind will be acting:
A = (l)(w) = (4.6 m)(1.5 m) = 6.9 m^2.

2. Use the area to find the rate of the wind in kg/s:
(66 kg/s/m^2)(6.9m^2) = 455.4 kg/s

3. Find the force of the wind:
We know that the force stopping the wind is exerted by the car, so the force on the car is equal in magnitude and opposite in direction to the force stopping the wind. So we can just look at magnitude to find the force of the wind:

Fwind on car = Fcar on wind = Δpwind/Δt = (mwind)(vwind)/Δt = (mwind/Δt)(vwind) = (455.4 kg/s)(42 m/s) = 19,127 N --> 19,000 N.

B)
Ffr = usmg = (0.40)(1500 kg)(9.8m/s^2) = 5880 N --> 5,900 N.

19,000 N is much greater than 5,900 N, meaning the force of the wind is strong enough to blow over a car.

Here's some evidence!

http://www.youtube.com/watch?v=F1H1EPwzf2c

Part 2: If the wind got the car to travel 10.0 m/s and it was blown into a large building (and came to rest), A. What is the velocity of the car/building after the collision? B. How much energy is "lost" in the collision?

A) We would use conservation of momentum to find the final velocity of the car/building:

mcvci + mbvbi = (mc + mb)vf

But in this case, we can assume that the mass of the building is so much greater than the mass of the car that the velocity of the building will be unchanged in the collision, meaning the velocity of the car/building after the collision is 0 m/s (since the building starts at rest).

B) To find the energy "lost" in the collision, we look at the change in kinetic energy of the system:

ΔKE = KEafter - KEbefore

Since the velocity of the car/building after the collision is zero, the kinetic energy of the system after the collision is also zero. This means ALL of the initial kinetic energy is "lost". But we know this just means it is transformed into other types of energy, such as heat and sound!


Catherine Stecyk

Tying an NFL Record

The physics of making a 63 yard field goal

By Alex Girden

™ http://www.youtube.com/watch?v=NfD25zFCxWg

What we Know

      Distance =63 yards

      1 yard=0.9144 meters

      Distance=58 meters

      Height of cross bar =10. feet

      1 foot=0.3048 meters

      Height =3.0 meters

      Football= 0.413 kg

      Estimated time of travel = 3 seconds

The easy idea

      Solving for initial velocity?

     Angle of kick?

     Kinetic Energy?

Calculations

      Δx=vot+1/2at^2

      3.o=vo (3 s)+ 0.5*-9.8* (3)^2

      Voy=16 m/s

      Max height- Vf^2=Vo^2+2aΔx

      Δx=13 meters

More equations

      Δx=vot+1/2at^2

      ax=0 m/s^2

      58=vo (3)

      Vox=19 m/s

      Vnet^2=vx^2 +vy^2

      Vnet=25 m/s

Unknowns

      sinΘ=16/25

      Θ=40 °

      Kinetic Energy given to ball

      KE=1/2mv^2

      ½ (0.413kg) (25)^2=129 J

Force of a kicker

      Time in contact with ball approximate=1.8 X 10^-3 s

      Vf=vo+at

      25=0 + a (1.3X 10^-3)

      Acceleration = 19,200 m/s

      Force=ma= (19,200 * 0.413 kg)=7,900 N

Future Directions

      Take into account wind speed and direction

      Air resistance’s affect on the ball

Physics of blocking a Puck

By Rachel Walsh

            For my physics new assignment I chose to evaluate the forces and momentum involved in stopping a puck with your forehead. This was inspired by this block by Ian Lapperriere http://www.youtube.com/watch?v=47ankjpg__Q . First I considered the change in momentum from before and after the hit and then considered the sum of the forces.

Momentum of puck before the block:

Average speed of a slap shot assumed to be 80 mph = 128.8 km/h = 35.8 m/s

  m = 1 pound or 0.454 kg

            p = mv = (35.8m/s)(0.454 kg) = 16.6 kg m/s

Momentum after the block:

v = - 25 m/s (estimated)

            p = mv = (-25 m/s)(0.454 kg) = - 11.35 kg m/s

Sum of forces:

 t of impact = 0.01 seconds

SF = Dp/Dt = (-11.35 kg m/s – 16.6 kg m/s)/(0.01s) = -2795 N

2795 N are applied by the player’s forehead to change the momentum of the puck. Due to Newton’s law stating that every force has an equal and opposite force the puck also applies 2795 N of force on Lapperiere’s forehead. In homework #3 it is stated that 6000 N is needed to fracture the human forehead; therefore, a puck would not be able to fracture the forehead in this example.

            

THE PHYSICS OF A LONG BACK SPINGING KICK

A long back spinning kick is a kick where the practitioner (me.) spins using the foot that is initially behind them to kick in a circle by raising their foot at they go. The kick is intended to hit the target 180° after motion has started.

* Please note this is actually not a circle it is an oval because I have to shift my weight as I go, but for this problem we’re going to call it a circle.

            I know that I can kick hard enough to break someone’s ribs. It takes 3300 newton’s to do this (the average martial arts master can, depending on the kick, get up to 9000 newton’s.) I’m not a master and I’m pretty small so let’s just stick to breaking a rib here. So torque is:

T=rFsinΘ

 Where Θ is 90. The length of my leg is about .69m and since I’m pivoting in a circle the distance to my foot is the radius. So the torque is going to be

.69*3300*sin90

Torque is 2277 Nm.

          Now to get the Radial Acceleration we use the equation T=Iα if we assume my leg is a thin straight object with the point that we rotate around being the end. Than the equation for torque should read. T=[(mL^2)/3]*α.

Now I have a mass of 50 kilograms. (well 49.8 but let’s call it 50.) And my leg to my hip joint is about .69m. And earlier we found the torque to be about 2300Nm so we get 

α=(T*3)/(ML^2)

α=(2300*3)/(50*.69^2).

α=23.8m/s^2

          So to get the angular velocity of my foot at the point of contact.

α=ω/t.

So I timed myself kicking and it takes me.75 seconds for a half a circle. Because we assume once I hit the guy I stop spinning.  

ω=αt

 23.8*.75=ω

 ω=17.9 m/s.

 So in order to get the linear velocity my foot hits the target with we take the angular velocity and times it by the radius.

V=ωr

17.9*.69=V

V=12.35m/s.

To give you a better reference for this number 12.35m/s is 27.63mph.

Physics of Dunking a Basketball

By Kathryn Taylor

Basketball players exert a huge amount of force when they are jumping and cutting during games. Dunking is one of the skills that requires the greatest amount of force to jump as high as possible and virtually drop the ball downwards through the hoop. In his day Michael Jordan was one of the greatest dunkers and was known as Air Jordan.

To make the situation a little simpler I am going to look at dunking a ball from just standing under the rim. This takes away the horizontal component of the situation which would have little effect anyway as there is negligible acceleration in that direction. Michael Jordan is 6’6” (1.98m) and in his playing days weighed 216lb (98kg) and would be dunking a ball of 624g. I found that he was thought to have a reach of 8’10” (2.7m) (http://thekitchensinkhole.blogspot.com/2007/02/

sinkhole-vertical-leaps.html). This means that the change in height for Michael Jordan dunking on a regulation rim (10’) would be just 1’2” (0.35m). This means that the minimum work done by his jump would be equal to ΔPE= mgΔh = (98+0.624kg)*(9.8m/sˆ2)*(0.35m) = 338.3J. This work was done over the distance that his knees bend which can be estimated to around 3’ (0.9144m). Using this the minimum applied force by Jordan into the ground to dunk the ball is equal to Work/distance = (338.3J/0.9144m)=370N.

The Guinness World Record dunk of 12ft was set by Michael “Wild Thing” Wilson who is part of the Harlem Globetrotters. For Jordan to complete this dunk his work would be mgΔh = (98.624kg)*(9.8m/s^2)*(0.96m) = 927.9J. This relates to an applied force = (927.9J/0.9144m)=1014.7N. This is a huge increase in force required for a dunk only 2 feet higher than regulation. 

Physics in Figure Skating: The Death Spiral

Figure skating is obviously an interesting application of physics, and the success of many of the "moves" that skaters do depends on whether or not they are able to exert the force that is physically required for what they are aiming to accomplish. Specifically, I chose to look at the physics behind the "death spiral," which is a common "move" that is is seeming difficult in figure skating. The death spiral is when a male skater pulls a female skater in a circle while she is almost completely perpendicular to the ice. See a video of it here: http://www.youtube.com/watch?v=bOsWG3BeslE

I chose to qualitatively look at the center of mass between the two skaters, simplifying their relationship to being one rigid body. Since CM=(m1x1+m2x2)/(m1+m2), I intuitively observe that the center of mass lies closer to the male skater. This makes sense because as a male skater who weighs more, his mass will be larger and therefore the resulting calculation for center of mass will lie closer to him.

Next I considered looking at the speed the pair would probably be skating in. Based on the video, I presumed that the pair did about one revolution in two seconds. To find the angular velocity, I know that w=2piT, so plugging in 0.5 for T estimates that the skaters' angular velocity was about 3.1 radians per second. Next, to find their approximate linear velocity I assumed their radius of rotation (from the center of mass to where the male plants his foot) was about 0.5 meters. Accordingly, I know that v=rw, so plugging in 0.5 in for r and 3.1 in for w, I found their linear velocity to be about 1.6 meters per second. 

Next, I chose to look at what forces are involved between this pair as they rotate. They have a centripetal acceleration that comes from the force that the male exerts on the female in planting his blade into the ice. This force is the centripetal force that keeps the pair rotating. Based upon Newton's second law, I know that F=ma. Furthermore, I know that radial acceleration is equal to w^2r. Plugging this in, I can get an equation to estimate the force, F=(m1+m2)w^2r. Supposing that the female weighed about 50 kg and the male about 79 kg, I estimate the force to be about 620 N.

In my interpretation, I thought it was important to consider whether this amount of force seems reasonable. The male must exert 620 N of force in order to successfully complete the "death spiral." In looking at the force that the male can exert with his body weight, I found that he could exert about 770 N (F=ma=79*9.8). Therefore, the 620 N that the male must exert is a reasonable estimate, and more importantly possible. This is obviously good news for the female skater, who must trust that the male skater can keep his skate stabilized while the force is exerted as she spins, otherwise the stunt could result in an accident. Lastly, another consideration in the calculation of this force is that the force of tension in the hands of the skaters must equal 620 N--the skaters grips must withstand the large force.

Based upon my investigations above, it is apparent that the death spiral is a pretty difficult stunt in the world of figure skating. This is probably why its completion receives high scores in the judging rounds of major competitions. 

Figures that may help visualize: 

Source: http://www.real-world-physics-problems.com/physics-of-figure-skating.html

The Physics of Hitting an Elk With You Car

By Heather Frank

This summer when I was on a road trip with my Dad in the Grand Canyon we hit an Elk with our car at 4:30 in the morning while driving in the dark at 45 mph. Although the car was totaled and the Elk died, my Dad and I were unharmed since we were wearing our seatbelts and the airbags deployed.

In order to analyze the forces that we felt during the accident I had to make a few estimations. I assumed that M_CAR=2000 kg, M_ELK=225 kg, V₁ of the car before the accident was 45 mph (20 m/s) and V₂ is the velocity of the car and elk after colliding. I treated this as an inelastic collision because after we hit the Elk we carried the Elk on the hood of our car for about .5 seconds. Then we traveled an additional 15m before coming to a stop. After finding the Velocity of the Car and the Elk after the collision (V₂) I was able to find the W_NC done on the car to make it stop by looking at the change in Kinetic Energy. This W_NC was a combination of the work done by the friction between the tires and the road and the work done by the car brakes to come to the stop. Finally, I looked at the Force exerted on my Dad and I during the accident and how that force was significantly decreased by us wearing seatbelts and the deployment of the airbags, which increased the time that we experienced a change in momentum. 

MV_CAR+MV_ElK=MV_{CAR AND ELK}

(2000)(20)+0=(2225)(V) Vfinal of Car and Elk= 18m/s

Then Elk fell off car and car traveled for additional 15 m.

Change in KE=W_NC=1/2m(v_f²-v₀²)

W_NC=1/2(2000)(0-18²)= -324000 J

This is combination of Ffr and Work done by brakes of car over 30m.

W_NC=Ffrd+Fbraked

324000=(.7)(2000)(9.8)(15)+ Fbrake(15)

Fbrake= 7880 N A LOT OF FORCE!

 Also it is important to note that the reason we did not sustain injury from the accident is because the Force exerted on us (change in momentum over time) was significantly decreased by the airbags and the seatbelts, by increasing the time in which we felt the change in momentum.

NOTE: These numbers are all estimates. I have a foggy memory of the accident and my times and distances might be slightly off but the ideas of Physics apply to this scenario.

PUMPKIN CHUNKIN’

 Pumpkin Chunkin'

By Michelle Bradley

It seems like it’s about that time in the semester where everyone starts getting overwhelmed by the amount of work they have. I know that when I have too much work I love taking TV breaks, which mostly consist of my favorite show “Modern Family”.  Recently, I was watching the Halloween and Thanksgiving episodes, and came across the scene where the entire family goes pumpkin chunkin’. I decided to take a look at these contests in real life, and saw that they nicely corresponded with our recent discussions about conservation of energy.

I thought it would be interesting to examine a scenario to find the velocity of the pumpkin right before it hits the ground.  I made the assumption that the rope they use on the show works the same way as a bungee cord would (there is no friction between it and the pumpkin) and there is no air resistance. Additionally, I set the mass of the pumpkin to be about 6 kg and the initial launch height to be 5 m.  Also, I added a F_applied of about 2205 N because people pull back on the cord (distance= 2m) so that the pumpkin may gain potential energy. Before the pumpkin is launched, it only contains potential energy, and by the time it launches all of that energy turns into kinetic. Therefore: 

                                ΔKE = -ΔPE + W_NC                                                  

W_NC= -F_appdcos180= F_appd

½ mv_f² =-(mgh_f – mgh_o) + F_appd                        

½(6 kg) v_f²= -(6 kg)(9.8 m/s²)(- 5m) + (2205)(2)

v_f = 39.60 m/s

The final velocity of the pumpkin in this scenario would be 39.60 m/s. If you wanted to increase the velocity, you could have a greater applied force when launching. When you have a great applied force it leads to a greater distance for the pumpkin to travel as well, which is after all what those judges are looking for! 

The Physics Behind the Red Bull Stratos Project

The Physics Behind the Red Bull Stratos Project

by Chelsea Gottschalk

http://www.youtube.com/watch?v=FHtvDA0W34I

            As mentioned in class and outlined in a NY Times article, Felix Baumgartner recently became the first man to break the sound barrier by gravity alone. Felix was taken to a vertical height of 128,100 ft (24 miles) by a hot air balloon, and then jumped down to the ground, using a special space-like suit and a parachute. Part of this mission, called the Red Bull Stratos project, was to test new spacesuits as well as escape tactics for extreme altitudes. Although the figures are not exact yet, it was estimated that Felix dropped when he was 128,100 ft above the ground. He fell for approximately 4 minutes (240 seconds), reaching a maximum speed of 833.9 mph (372.787 m/s) before deploying a parachute about 1 mile (1609.3 m) above a New Mexico desert.

         Let’s take a look at the physics behind Felix’s super stunt.

Let’s assume that Felix has a mass of approximately 75 Kg and his space suit weighs around 70lbs (31.75Kg) (around what some models weigh), therefore: Total mass= 106.75 Kg

Also, since Felix will be constantly accelerating, the maximum speed will be equal to the point just before he opens his chute

There are two reference frames that we should assess:

1) The descent from the highest point to right when Felix deploys his parachute (this represents the total time under complete free fall) (37,435.5 m)

2) The descent from the point Felix opens his parachute to the time he touches the ground (1609.3 m)

Let’s look at the free fall first:

We know that Wnet=Wc+Wnc, where the Wnc represents the air resistance encountered from the fall

From the law of conservation of energy:

KEi + PEi = KEf + PEf – Wnc

1/2mv^2 + mgh = 1/2mv^2 + mgh –Wnc

Vi= 0 m/s                hi= 37,435.5 m

Vf= 372.787 m/s     hf = 0 m

0 + (106.75)(9.8)(37,435.5) = (1/2)(106.75)(372.787)^2 – Wnc

Wnc = -3.17 x 10^7 J of air resistance

F x d = -3.17 x 10^7 J   d=37,435.5 m

Fair= 846.79 = 847 N of air resistance opposing the direction of motion

Now, let’s look at what Felix’s max speed would be if there were no nonconserved forces:

Vf ^2 = Vi^2 + 2ay

Vf= sqrt (9.8 x 37435.5)

Vf= 856.58 = 857 m/s, over 2x as fast!

The speed of sound is 340.29 m/s…..

Now, let’s look at the second reference frame, from the moment Felix opens his parachute until he reaches the ground, covering a height of 1 mile, or 1609.3 m

We can use the same equation as before to find the force of the drag produced by the parachute:

KEi + PEi = KEf + PEf – Wnc

1/2mv^2 + mgh = 1/2mv^2 + mgh –Wnc

Vi = 372.787 m/s    hi= 1609.3 m

Vf= 0 m/s                hf= 0 m

(106.75)g(1609.3) + (1/2)(106.75)(372.787)^2 = 0 + 0 – Wnc

Wnc= -9.1 x 10^6 J

F x d = -9.1 x 10^6 J   d=1609.3 m

Fparachute= 5654.63 = 5650 N of force in the direction opposing motion- that is a lot!

Let’s look at the magnitude of the deceleration to see how it compares to the limits of the human body (5g’s before loss of consciousness):

Vf ^2 = Vi^2 + 2ay

0 = 372.787^2 + 2(1609.3)a

a= -43.1772 = -43.2 m/s^2 / 9.8 = 4.4g’s

It looks like Felix was very close to approaching the limit of the human body, but he survived, no doubt due to the careful physical calculations of this awesome stunt!

References: http://www.nytimes.com/2012/10/15/us/felix-baumgartner-skydiving.html?_r=0

The Physics of Frogs and Muscles

With our liberal arts education here at Colgate, I figured exploring the cross over between my Human Physiology class and our Physics class might be a little fun. In Human Physiology lab a little while ago, we performed an experiment on the gastrocnemius muscle of frogs (This more or less translates to the calf muscle in the back of the lower leg for both us and frogs). In our experiment, we prepped the isolated muscle and set it up to perform isotonic muscle twitches with varying weights. By finding the distance the muscle contracts and knowing the weight, we can easily calculate the work done by the muscle to lift the weights. Then, by plotting a graph of work done for a given weight lifted, I was able to obtain an optimal load at which the muscle does the greatest amount of work.

Accounting for the forces in play, we know that the force of gravity (mg) is determined by the varying weights we add to the muscle. Furthermore, we know that a force applied comes into play when the muscle contracts (and does work) to lift the weight. In order to simplify calculations, we ignored other forces besides the force of gravity on the weights and force applied. We ignored the weight of the muscle, and focused solely on the work done by the muscle on the weights. Our equation W = F * d helps us find the work done by the muscle on the weight.

My raw data for the lab was:

Weight (g)	Work	Height Lifted (mm)	Height (mv)
10	3.52725	0.352725	4.703
9	3.4803	0.3867	5.156
8	9.7782	1.222275	16.297
7	13.86368	1.980525	26.407
6	22.56345	3.760575	50.141
5	24.04088	4.808175	64.109
4	21.9609	5.490225	73.203
3	18.39974	6.133245	81.7766
2	13.95465	6.977325	93.031
1	7.6137	7.6137	101.516

Since we were stimulating the muscle at varying voltages, and similarly measuring the peak of the contraction in mV, I had to use a conversion factor to obtain the shortening distance (height lifted) for the muscle. This conversion factor was given as 75 mm / V. Also, to keep calculations simple in lab, we initially ignored gravity (as it multiples every mass by 9.8 equally), and considered the force of gravity = m (weight). This means that the force of contraction must overcome that downward force in order to have the weights move upward, and thus do work. To make this all a little more clear, I converted everything in SI units and corrected the force downward (and thus minimum upward) by multiplying gravity in.

Weight (kg)	Min. Force Exerted Upwards (N)	Distance (m)	Work Done (J)
0.01	0.098	3.53E-04	3.46E-05
0.009	0.0882	3.87E-04	3.41E-05
0.008	0.0784	1.22E-03	9.58E-05
0.007	0.0686	1.98E-03	1.36E-04
0.006	0.0588	3.76E-03	2.21E-04
0.005	0.049	4.81E-03	2.36E-04
0.004	0.0392	5.49E-03	2.15E-04
0.003	0.0294	6.13E-03	1.80E-04
0.002	0.0196	6.98E-03	1.37E-04
0.001	0.0098	7.61E-03	7.46E-05

Now, having everything in SI, I plotted the work vs. weight graph to find the optimal load where this muscle does the greatest amount of work using the numbers above, and found that this frog's gastrocnemius muscle's optimal load was 0.005kg, or 5g.


Let's take this a step further and calculate the power of this muscle.
Knowing power is the rate at which work is done, to find our average power of this muscle we can use the equation:
P = W / t.
For our isotonic muscle twitches, a contraction cycle lasted approx. 2 ms (variations occurred with changing weights, changing stimuli strength, how fatigued the muscle was, etc., but for simplicity's sake, 2 ms is a reasonable assumption).

At an optimal load of 5g (0.005kg), our muscle did 2.36E-4 Joules of work. Since 2 ms is 0.002 seconds, we simply divide to find that the frog's gastrocnemius muscle had an average power output of 0.118 W (J/s).

That's nice and all, but since I like to be self-centered now and then, how does that relate to us (and since the course is Human Physiology, how does it relate to humans?) Doing a little bit of research, I found the optimal power outputs of human gastrocnemius muscles at a walk and at a run in this article:
http://www.pnas.org/content/early/2012/01/04/1107972109.full.pdf

Granted they do some very interesting research and measure a variety of things, including changes in length for different parts of the muscle and power among different groupings, I used their data for muscle-tendon units to compare to my frog's gastrocnemius muscle power.

At a walk, the power output is around 37 W, and while running, our power output increases to around 48W.

Where frogs have a gastrocnemius muscle composed of fast twitch glycolytic muscle fibers that are useful for jumping quickly and moving only their small bodies, a power output of 0.118W makes sense. On the other hand, human gastrocnemius muscles contain a variety of fast twitch glyocolytic, fast twitch glycolytic oxidative, and slow twitch oxidative fibers. From an evolution standpoint, we need these variety of fibers to enable us to quickly run and jump (explosive movements ~ fast twitch fibers), but also maintain a standing posture throughout the day (long, gradual movements ~ slow twitch fibers). In order to maintain the work throughout the day, and to move our much larger masses, it makes sense that our power outputs are much higher than that of a frog's.







Farris DJ, Sawicki GS. Human medial gastrocnemius force-velocity behavior shifts
     with locomotion speed and gait. Proc Natl Acad Sci U S A. 2012;109:977-982. Found at  
     <http://www.pnas.org/content/early/2012/01/04/1107972109.full.pdf>.