It takes a flea \(1.0 \times 10^{-3}\) s to reach a peak speed of $0.74 \mathrm{m} / \mathrm{s}$ (a) If the mass of the flea is \(0.45 \times 10^{-6} \mathrm{kg},\) what is the average power required? (b) Insect muscle has a maximum output of 60 W/kg. If \(20 \%\) of the flea's weight is muscle, can the muscle provide the power needed? (c) The flea has a resilin pad at the base of the hind leg that compresses when the flea bends its leg to jump. If we assume the pad is a cube with a side of \(6.0 \times 10^{-5} \mathrm{m},\) and the pad compresses fully, what is the energy stored in the compression of the pads of the two hind legs? The Young's modulus for resilin is $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} .$ (d) Does this provide enough power for the jump?

We are given the flea's peak speed \(v = 0.74 \mathrm{m} / \mathrm{s}\) and the time it takes to reach that speed, \(t = 1.0 \times 10^{-3} \mathrm{s}\). The average acceleration can be calculated using the formula \(a = \frac{v}{t}\). $$a = \frac{0.74 \mathrm{m/s}}{1.0 \times 10^{-3} \mathrm{s}} = 740 \mathrm{m/s^2}$$

Now, we will calculate the force required to achieve the peak speed using the formula \(F = ma\), where \(m = 0.45 \times 10^{-6} \mathrm{kg}\) is the mass of the flea and \(a = 740 \mathrm{m/s^2}\) is the average acceleration calculated in step 1. $$F = (0.45 \times 10^{-6} \mathrm{kg})(740 \mathrm{m/s^2}) = 3.33\times 10^{-4} \mathrm{N}$$

The formula to calculate the average power is \(P = Fv\), where \(F = 3.33\times 10^{-4} \mathrm{N}\) is the force and \(v = 0.74 \mathrm{m/s}\) is the peak speed. $$P = (3.33\times 10^{-4} \mathrm{N})(0.74 \mathrm{m/s}) = 2.46\times 10^{-4} \mathrm{W}$$

Insect muscle has a maximum output of 60 W/kg. The flea's muscle weight is \(20\%\) of the total mass, so \(m_\text{muscle} = 0.20(0.45 \times 10^{-6} \mathrm{kg}) = 0.09 \times 10^{-6}\mathrm{kg}\). The power output of the flea's muscle is given by \((60 \mathrm{W/kg})(0.09 \times 10^{-6} \mathrm{kg}) = 5.4 \times 10^{-6} \mathrm{W}\). Since the needed power is \(2.46\times 10^{-4} \mathrm{W}\), the muscle can provide the power needed as \(2.46\times 10^{-4} \mathrm{W} < 5.4 \times 10^{-6} \mathrm{W}\).

The formula to calculate the potential energy stored in a compressed spring is \(U = \frac{1}{2}kx^2\), where \(k\) is the spring constant and \(x\) is the compression distance. We are given that the resilin pad is a cube with a side of length \(6.0 \times 10^{-5} \mathrm{m}\). The volume of the pad is \(V = (6.0 \times 10^{-5} \mathrm{m})^3\). The compression distance \(x\) is equal to the side length of the cube. The spring constant can be related to the Young's modulus, \(Y\), with the formula \(k = (\frac{Y}{V})x^2\). The Young's modulus of resilin is given as \(Y = 1.7 \times 10^{6} \mathrm{N/m^2}\). Using these values, we can calculate \(k\): $$k = (\frac{1.7 \times 10^{6} \mathrm{N/m^2}}{V})x^2 = \frac{(1.7 \times 10^{6} \mathrm{N/m^2}) (6.0 \times 10^{-5} \mathrm{m})^2}{(6.0 \times 10^{-5} \mathrm{m})^3}$$ Now, calculate the potential energy stored in one pad: $$U_\text{pad} = \frac{1}{2}kx^2$$ Since there are two pads, the total energy stored is: $$U_\text{total} = 2U_\text{pad}$$

The stored energy, \(U_\text{total}\), must be compared with the required instantaneous power for the jump, \(P_{req} = 2.46\times 10^{-4} \mathrm{W}\), remembering that the power is work done per unit time (\(P = \frac{W}{t}\)). If the energy is enough for the jump, we should have \(U_\text{total} \geq P_{req}t\). If the stored energy is greater or equal to the required power, the resilin pads can provide enough power for the jump. If the stored energy is less than the required power, the resilin pads do not provide enough power for the jump.