Answer
(a) The kinetic energy of the locust is 0.014 J.
(b) The locust required 0.04 J of energy for the jump.
Work Step by Step
(a) $KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(0.0030 ~kg)(3.0 ~m/s)^2$
$KE = 0.014 ~J$
The kinetic energy of the locust is 0.014 J.
(b) $(0.35)(E) = 0.014 ~J$
$E = \frac{0.014 ~J}{0.35}$
$E = 0.04 ~J$
The locust required 0.04 J of energy for the jump.
