Q43P - University Physics with Modern Physics

Question

The froghopper (Philaenus spumarius), the champion leaper of the insect world, has a mass of  and leaves the ground (in the most energetic jumps) at 4.0 m/s from a vertical start. The jump itself lasts a mere 1.0 ms before the insect is clear of the ground. Assuming constant acceleration, (a) draw a free-body diagram of this mighty leaper during the jump; (b) find the force that the ground exerts on the froghopper during the jump; and (c) express the force in part (b) in terms of the froghopper’s weight.

Step-by-Step Solution

Answer

(a) .

(b) The force that the ground exerts on the froghopper during the jump is $4.9\times {10}^{-2}\mathrm{N}$.

(c) The force in terms of the froghoppers’ weight is 409 mg.

Step 1: Identification of the given data

 

The given data is listed below as:

 

• The mass of the froghopper is, $\mathrm{m}=12.3 \mathrm{mg}\times \left(\frac{{10}^{-6} \mathrm{kg}}{1 \mathrm{mg}}\right)=12.3\times {10}^{-6} \mathrm{kg}$
• The final velocity froghopper is, $\mathrm{v}=4.0 \mathrm{m}/\mathrm{s}$
• The time taken for the jump is, $\mathrm{t}=1 \mathrm{ms}\times \left(\frac{{10}^{-3}\mathrm{s} }{1 \mathrm{ms}}\right)=1\times {10}^{-3} \mathrm{s}$

Step 2: Significance of the Newton’s second law

The second law states that the force of an object is directly proportional to the product of the mass and the acceleration of an object. The law mainly gives the acceleration of a point particle.

Step 3: (a) Determination of the free-body diagram of the leaper

The free body diagram of the leaper has been drawn below:

In the above diagram, it has been identified that the weight of the leaper w is acting downwards that is the product of the mass m and the acceleration due to gravity g. The normal force acting above the ball is $F_N$. The acceleration of the ball $a_{av,y}$ is directed upwards.

Step 4: (b) Determination of the force that the ground exerts on the froghopper

The equation of the acceleration of the froghopper is expressed as:

 

$$\begin{gathered} v=u+at \\ a=\frac{v-u}{t} \end{gathered}$$

 

Here, v is the final speed, u is the initial speed, a is the acceleration of the froghopper and t is the time taken for the jump.

 

The equation of the force exerted is expressed as:

 

$$\begin{gathered} F=mg+ma \\ =mg+m\frac{v-u}{t} \end{gathered}$$ 

 

Here, m is the mass of the froghopper and g is the acceleration due to gravity.

 

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$ for g, $12.3\times {10}^{-6} \mathrm{kg}$ for m , 0 for u, $4.0m/s$ for v and $1\times {10}^{-3} s$ for t in the above equation.

$$\begin{gathered} \mathrm{F}=\left(12.3\times {10}^{-6} \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)+\left(12.3\times {10}^{-6} \mathrm{kg}\right)\frac{\left(4.0 \mathrm{m}/\mathrm{s}\right)-0}{\left(1\times {10}^{-3}\right)} \\ =1.2054\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+\left(12.3\times {10}^{-6} \mathrm{kg}\right)\left(4000\mathrm{m}/{\mathrm{s}}^2\right) \\ =1.2054\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2+0.0492 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ = 4.9\times {10}^{-2} \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Hence, further as:

$$\begin{gathered} \mathrm{F}=4.9\times {10}^{-2} \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =4.9\times {10}^{-2} \mathrm{N} \end{gathered}$$ 

 

 

Thus, the force that the ground exerts on the froghopper during the jump is $4.9\times {10}^{-2} \mathrm{N}$.

Step 5: (c) Determination of the force in terms of the weight

The equation of the force in terms of the weight is expressed as:

$$\begin{gathered} F=W+\frac{a}{g} \\ F=W+\frac{v-u}{tg} \end{gathered}$$ 

 

 

Here, W is the weight of the froghopper which is equal to mg and a is the acceleration due to gravity.

 

Substitute the values in the above equation.

$$\begin{gathered} F=W+\frac{\left(4.0 m/s\right)-0}{\left(1\times {10}^{-3} s\right)\left(9.8 m/s^2\right)} \\ =mg+\frac{\left(4.0 m/s\right)}{\left(9.8\times {10}^{-3} m/s\right)} \\ =mg+408 \\ =409 mg \end{gathered}$$ 

 

Thus, the force in terms of the froghoppers’ weight is 409mg.