Question

An average froghopper insect has a mass of 12.8 mg and jumps to a maximum height of 293 mm when its takeoff angle is 62.0∘ above the horizontal. With the takeoff speed being 2.71 m/s :

a) How much kinetic energy did the froghopper generate for this jump? Express your answer in microjoules.
b) How much energy per unit body mass was required for the jump? Express your answer in joules per kilogram of body mass.

Answer 1

Step-by-step explanation:

a) The potential energy gained by the froghopper at the maximum height of 293 mm can be calculated using the formula:

ΔPE = mgh

where ΔPE is the change in potential energy, m is the mass, g is the acceleration due to gravity, and h is the maximum height.

Substituting the given values, we get:

ΔPE = (12.8 × 10^-6 kg) × (9.81 m/s^2) × (0.293 m) = 3.69 × 10^-6 J

The kinetic energy of the froghopper at takeoff can be calculated using the formula:

KE = 0.5mv^2

where KE is the kinetic energy and v is the takeoff speed.

Substituting the given values, we get:

KE = 0.5 × (12.8 × 10^-6 kg) × (2.71 m/s)^2 = 4.75 × 10^-5 J

Therefore, the total energy generated by the froghopper for the jump is the sum of the potential and kinetic energy, which is:

Total energy = ΔPE + KE = 3.69 × 10^-6 J + 4.75 × 10^-5 J = 5.12 × 10^-5 J

Expressing the answer in microjoules, we get:

Total energy = 5.12 × 10^-5 J = 51.2 µJ

b) The energy per unit body mass required for the jump can be calculated by dividing the total energy generated by the froghopper by its body mass.

Substituting the given values, we get:

Energy per unit body mass = (5.12 × 10^-5 J) ÷ (12.8 × 10^-6 kg) = 4 J/kg

Therefore, the energy per unit body mass required for the jump is 4 J/kg.