(a) A 3.0-g locust reaches a speed of 3.0 ms during its jump. What is its kinetic energy at this speed? (b) If the locust transforms energy with 35% efficiency, how much energy is required for the jump?

(a) The kinetic energy of the locust is 0.027 J.

(b) The energy required for the locust to jump is 0.00945 J.

Kinetic energy is the energy possessed by a moving body.

The formula of kinetic energy is

K.E = mv²/2 Equation 1

Where K.E = Kinetic energy, m = mass, v = velocity.

(a) From the question,

Given: m = 3 g = 0.003 kg, v = 3.0 m/s

Substitute these values into equation 1

K.E = 0.003(3²)/2

K.E = 0.027 J

(b) From the question,

The energy required for the locust to jump is

E' = 0.35(K.E) Equation 2

substituting the value of K.E into equation 2

E' = 0.35(0.027)

E' = 0.00945 J

Hence, (a) The kinetic energy of the locust is 0.027 J (b) The energy required for the locust to jump is 0.00945 J.

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(a). Kinetic energy = 0.0135J

(b). E_{in} = 0.039J

Explanation:

(a).

The kinetic energy of the locust is

(b).

A 35% efficiency means that 35% of the input energy is converted into useful mechanical energy; therefore, if input energy is , 35% of appears as the kinetic energy of the locust:

which the amount of energy required for the jump.

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