Question

A 0.10 g flea, having leapt from the surface of a dog’s cranium, is observed to be moving at 1.25 m/s when it is 5.00 cm above the position from which it leapt. What was the elastic potential energy stored in its legs before its leap

Answer 1

Answer:

E.P.E = 1.27 x 10⁻⁴ J = 0.127 mJ

Explanation:

From Law of Conservation Energy, we know that:

Elastic Potential Energy Stored = Gain in Kinetic Energy + Gain in Gravitational Potential Energy

E.P.E = (1/2)mv² + mgh

where,

E.P.E = Elastic Potential Energy = ?

m = mass of flea = 0.1 g = 1 x 10⁻⁴ kg

v = speed = 1.25 m/s

g = 9.8 m/s²

h = height = 5 cm = 0.05 m

Therefore,

E.P.E = (1/2)(1 x 10⁻⁴ kg)(1.25 m/s)² + (1 x 10⁻⁴ kg)(9.8 m/s²)(0.05 m)

E.P.E = 0.78 x 10⁻⁴ J + 0.49 x 10⁻⁴ J

E.P.E = 1.27 x 10⁻⁴ J = 0.127 mJ