Molly eats a 500 kcal (2.09x106 J power bar before the big pole vault. The bar?s

Question

Molly eats a 500 kcal (2.09x106 J power bar before the big pole vault. The bar?s energy content comes from changing chemical bonds from a high to a low state and expelling gases. However, 25.0% of the bar?s energy is lost expelling gases and 60.0% is needed by the body for various biological functions. How much energy is available to Molly for the run? Energy losses due to air resistance and friction on the run are 200,000 J, Molly?s increased heart rate and blood pressure use 55,000 J of the available energy during the run. What top speed can the 50.0 kg Molly expect to attain? The kinetic energy is transferred to the pole, which is compressed like a spring of elastic constant of 2720 N/m; Air resistance energy loss on the way up is 300 J, and as she crosses the bar she has a horizontal speed of 2.00 m/s. If Molly rises to a height equal to the expansion of the pole what is that height she reaches? On the way down she encounters another 300 J of air resistance. How much heat in the end is given up when she hits the dirt and comes to a stop The problem requires us to break down sequences into several stages. I think energy conservation should be considered here. This thread is the start for our discussion to solve the problem, feel free to contribute your ideal/approach.

Explanation / Answer

a))

Energy lost = 25 percent

energy used for biological functions = 60 percent

So, remaining precent = 100 - 25 -60 = 15 percent

So, energy available to Molly = 15 percent of 500 kcal

= (15/100)*500 = 75 kcal = 75*4.18

= 313.5 kJ <-------answer

b)

Energy losses due to run = 200 kJ

due to increased heart rate = 55 kJ

So, remaining energy = 313.5 - 200 - 55 = 58.5 kJ

The Kinetic energy = 58.5 kJ

So, 0.5*mv^2 = 58.5 kJ

where v = top speed

So, 0.5*50*v^2 = 58.5*10^3

So, v = 48.4 m/s <--------answer

c)

Now, the Kinetic energy is transfered to the spring

So, Kinetic energy = spring potential energy

So, KE = PE

So, 0.5*mv^2 = 0.5*kx^2 = 58.5*10^3

where k = spring constant = 2720 N/m

x = expansion of spring

So, 0.5*2720*x^2 = 58.5*10^3

So, x = 6.56 m <------------answer(height reached)

Now, as he reaches max height,

the gravitational Potential energy gained, PEg = mgx = 50*9.8*6.56 = 3214.4 J = 3.214 kJ

Total Energy lost to air resistance = 2*300 J = 0.6 kJ <------ on the eay up and then down

So, remaining energy,= 58.5 - 0.6 = 57.9 kJ <---------answer (heat given up when he hits the dirt)

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