BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.0-mm-long, 0.50-mg flea can reach a height of 20 cm in a single leap. (a) Ignoring air drag, what is the take-off speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a 65-kg, 2.0-m-tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would the man need? (d) Most humans can jump no more than 60 cm from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 65-kg person? (e) Where does the flea store the energy that allows it to make sudden leaps?
Solution 7E Step 1: a) suppose, the initial velocity of the flea as “u”. The final velocity of the flea will be zero at a height of 20 cm. There will be an influence of acceleration due to gravity on the flea and it will have a negative sign since it is flying up. Therefore, we can use the Newton’s equations of motion, v = u + 2aS here. 2 2 S - distance of travel. Here, S = 20 cm = 0.2 m That is, v = u - 2gS The final velocity of the flea, v = 0 m/s 2 2 Therefore, 0 m/s = u - 2 × 9.8 m/s × 0.2 m 0 m/s = u - 3.92 m /s 2 2 2 2 2 Rearranging the equation, we get, u = 3.92 m /s Taking square roots on both sides we get, u = 1.98 m/s The take-off velocity of the flea will be, u = 1.98 m/s Step 2: 2 b) Kinetic energy of the flea during take off, KE take-off½ mu Mass of the flea, m = 0.50 mg -3 1 mg = 10 g Therefore, m = 0.50 × 10 g -3 -3 1 g = 10 kg Therefore, m = 0.50 × 10 × 10 kg -3 m = 0.50 × 10 kg -6 -6 2 2 2 Therefore, KE take-off½ ×0.50 × 10 kg × 1.98 m /s KE take-off0.98 × 10 J -6 Kinetic energy per kilogram of mass,KE / mass of flea = 0.98 × 10 J / 0.50 × 10 -6 take-off kg KE take-offmass of flea = 1.96 J/kg Step 3: c) ratio of height of jump of the flea with the flea’s size, R = 20 cm / 2 mm = 20 cm / 0.2 cm R = 100 If a man with height 2 m is jumping by obeying this ratio, then, R = height of jump / height of man 100 = Height of jump / 2 m Then, height of jump = 200 m The take off speed needed can be calculated by using the equation, v = u - 2gS 2 2 2 2 0 = u - 2×9.8 m/s × 200 m Rearranging the equation we get, u = 3920 m /s2 2 2 Taking square root on both sides, u = 62.61 m/s The take-off velocity of the human to cover this height would be, u = 62.61 m/s