Question

Fleas have remarkable jumping ability. A 0.45 mg flea, jumping straight up, would reach a height of 30 cm if there were no air resistance. In reality, air resistance limits the height to 20 cm .
Part A What is the flea's kinetic energy as it leaves the ground?
Part B At its highest point, what fraction of the initial kinetic energy has been converted to potential energy?Express your answer using two significant figures.

Answer 1

Answer:

A) $1.35*10^{-6}J$

B) 0.67

Explanation:

If there would be no air resistance:

$Vf^2=Vo^2-2*g*h$

$0=Vo^2-2*10*0.3$

$Vo=\sqrt{6} m/s$

So, its initail kinetic energy is:

$Ko = m/2*Vo^2=0.45/10^6/2*6=1.35*10^{-6}J$

Its final potential energy is:

$Uf = m*g*h = 0.45/10^6*10*0.2=0.9*10^{-6}J$

The fraction of the potential energy is:

Fraction = Uf/Ko = 0.67

This means that one third of its initial energy is lost.

Answer 2

The theorem of conservation of mechanical energy allows to find the results for the initial energy and the energy loss due to friction are:

  A) the initial kinetic energy is:      K = 1.33 10⁻⁶ J

  B) The fraction of energy dissipated is:       $\frac{U}{K}$  = 0.66

The theorem for the conservation of mechanical energy states that energy cannot be created or destroyed only it can be transformed.

Mechanical energy is defined as the sum of kinetic energy plus potential energies.

               Em = K + U

Let's find the speed at which the flea jumps if there were no friction in the air. In this case we write the mechanical energy at two points

Initial point.

      Emo = K = ½ m v²

Final point. At the highest height y = 0.30 m because there is no friction

      Em_f = U = mg h

Energy is conserved

      Em₀ = Em_f

      ½ m v² = m g h

      v = $\sqrt{2gh}$  

      v = $\sqrt{2 \ 9.8 \ 0.30}$  

      v = 2.43 m / s

A) They ask for the initial kinetic energy

      K = ½ m v²

      K = ½ 0.45 10⁻⁶ 2.43²

      K = 1.33 10⁻⁶ J

B) They ask for the fraction of energy lost if we take into account the friction of the air

let's find the initial kinetic energy

    K = ½ m v²

     

We look for the potential energy until h’= 20 cm = 0.20 m that manages to jump

    U = m g h '

The fraction of energy lost is

     $\frac{U}{K} = \frac{mgh' }{\frac{1}{2} m v^2 }$  

     $\frac{U}{K} = \frac{2gh' }{v^2 }$

Let's calculate  

     $\frac{U}{K} = \frac{2 \ 9.8 \ 0.20 }{2.43^2 }$  

     $\frac{U}{K}$  = 0.66

In conclusion, using the theorem of conservation of mechanical energy we can find the results for the initial energy and the energy loss due to friction are:

  A) the initial kinetic energy is:    K = 1.33 10⁻⁶ J

  B) The fraction of energy dissipated is:       $\frac{U}{K}$  = 0.66

Learn more here: brainly.com/question/14688403