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.

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

Answer 1

a) The takeoff speed of the froghopper is 1.4 m/s. b) The froghopper generated 14 microjoules of kinetic energy for this jump. c) The energy per unit body mass required for this jump is 1.09 joules per kilogram of body mass.

Step-by-step explanation:

a) Finding the takeoff speed:

We can use the principle of conservation of energy to find the takeoff speed of the froghopper. At the maximum height of the jump, all of the initial kinetic energy will be converted to gravitational potential energy. So:

Initial kinetic energy = Final potential energy

0.5 * m * (vx)2 = m * g * h

Solving for vx:

(vx)2 = 2 * g * h

vx = √(2 * g * h)

Substituting the given values:

vx = √(2 * 9.8 * 0.293) = 1.4 m/s

Therefore, the takeoff speed of the froghopper is 1.4 m/s.

b) Finding the kinetic energy generated:

The kinetic energy can be calculated using the formula:

Kinetic energy = 0.5 * m * (vx)2

Substituting the given values:

Kinetic energy = 0.5 * 0.0128 * (1.4)2 = 0.014 joules = 14 microjoules

Therefore, the froghopper generated 14 microjoules of kinetic energy for this jump.

c) Finding the energy per unit body mass:

The energy per unit body mass can be calculated using the formula:

Energy per unit body mass = Kinetic energy / Mass = 0.014 / 0.0128 = 1.09 joules per kilogram of body mass

Therefore, the energy per unit body mass required for this jump is 1.09 joules per kilogram of body mass.

Answer 2

a) The takeoff speed of the froghopper can be found using the following equation:

v^2 = 2gh/(1 - cos^2(theta))

where:

v = takeoff speed

g = acceleration due to gravity (9.81 m/s^2)

h = maximum height (293 mm = 0.293 m)

theta = takeoff angle (62.0 degrees)

Substituting the given values into the equation, we get:

v^2 = 2(9.81)(0.293)/(1 - cos^2(62.0))

v^2 = 0.571

v = sqrt(0.571)

v ≈ 0.756 m/s

Therefore, the takeoff speed of the froghopper is approximately 0.756 m/s.

b) The kinetic energy generated by the froghopper can be found using the following equation:

KE = 0.5mv^2

where:

m = mass (12.8 mg = 0.0128 g)

v = takeoff speed (0.756 m/s)

Substituting the given values into the equation, we get:

KE = 0.5(0.0128)(0.756)^2

KE ≈ 0.00346 J

(1 J = 10^6 microjoules)

Therefore, the kinetic energy generated by the froghopper for this jump is approximately 0.00346 microjoules.

c) The energy per unit body mass required for this jump can be found by dividing the kinetic energy by the mass of the froghopper:

energy per unit body mass = KE/m

Substituting the values we obtained earlier, we get:

energy per unit body mass = 0.00346/0.0128

energy per unit body mass ≈ 0.270 J/kg

Therefore, the energy per unit body mass required for this jump is approximately 0.270 joules per kilogram of body mass.