Answer
(a) 2.0 m/s
(b) 2.0 J/kg
(c) 63 m/s
(d) 5.9 J/kg
Work Step by Step
Remember: KE (kinetic energy) is converted to GPE (gravitational potential energy) when the flea jumps.
Thus, we can put $m\times g \times h$ = $\frac{1}{2}\times mv^{2}$. With a bit of rearrangement and algebraic manipulation, it becomes $v$ = $\sqrt {2gh}$, where $g$ = 10.
(a) Since the critter can reach a height of 20 cm in a single leap, we can plug it into the formula.
$v$ = $\sqrt {2gh}$ = $\sqrt {2 \times (9.80 m/s )(0.20 m)}$ = 2.0 m/s.
Therefore, the critter has a take-off speed of 2.0 m/s.
(b) KE = $\frac{1}{2} \times m \times v^{2}$, but KE also equals $m\times g\times h$. So, plugging in the mass, which is 0.50-mg, or $(0.50\times 10^{-6}) kg $ and the take-off speed we calculated above, of 2.0 m/s, we can work out the kinetic energy first.
$K_{i}$ = $m\times g\times h$ = $(0.50 \times 10^{-6} kg)(9.80 m/s )(0.20 m)$ = $9.8 \times 10^{-7}$ J.
The kinetic energy per kilogram is simply the kinetic energy calculated above divided by the mass in kilograms. Therefore:
$\frac{K_{i}}{m}$ = $\frac{9.8 \times 10^{-7} J}{0.50 \times 10^{-6} kg}$ = 2.0 $J/kg$.
(c) It is stated that the ratio of human jump height to human length is the same as critter jump height to critter length. So, we can easily calculate this:
$h_{h} = h_{f}\times (\frac{c_{h}}{c_{f}})(0.20 m)(\frac{2.0 m}{2.0\times 10^{-3} m}) = 200 m$
$v_{h} = \sqrt (2\times g \times h) = \sqrt (2\times (9.80 m/s^{2}) \times (200 m) = 63 m/s$
(d) As above, human kinetic energy per kilogram is easily calculated by dividing total KE by mass:
$\frac{K_{i}}{m} = g\times h = (9.80 m/s^{2}) \times (0.60 m) = 5.9 J/kg$
