Question

The froghopper (philaenus spumarius, the champion leaper of the insect world, has a mass of 12.3mg and leaves the ground (in the most energetic jumps at 4.00m/s from a vertical start. the jump itself lasts a mere 1.00ms before the insect is clear of the ground. (a find the force that the ground exerts on the froghopper during its jump. (b express the force of (a in terms of the frog hoppers weight.

Answer 1

First, the acceleration is solved by dividing the velocity by time.
                                  acceleration = 4 m/s / (1x10^-3s) = 4000 m/s²
Then, we solve for the force by multiplying the mass by acceleration.
                                        F = m x a
Substituting the known values,
                                      F = (12.3 mg)(4000 m/s²) = (12.3 x 10^-6 kg)(4000 m/s²)
                                          F = 0.0492 N
Thus, the force of the froghopper is approximately equal to 0.0492 N. 

Answer 2

Explanation :

It is given that :

Mass of frog hopper, $m=12.3\ mg=12.3\times 10^{-6}\ kg$

Velocity of frog hopper, v = 4 m/s

The time duration of a jump, $t=1\ ms=10^{-3}\ s$

(a) We know that force, F = m a

and acceleration is rate of change of velocity.

So, $F=m\dfrac{v}{t}$

Putting all the values in above equation we get:

$F=12.3\times 10^{-6}\ kg\times\dfrac{4\ m/s}{10^{-3}\ s}$

$F=49.2\times 10^{-3}\ N$

$F=0.0492\ N$

(b) Force in terms of the frog hoppers weight is given by : $\dfrac{F}{mg}$

$\dfrac{0.0492\ N}{12.3\times 10^{-6}\ kg\times 9.8\ m/s^2}=408\ N$

Hence, this is the required solution.