Chapter 4, Problem 91QAP

To determine

(a)

The time of the jump and acceleration during the time.

Answer to Problem 91QAP

The time of the jump is $1.38\text{ms}$ and acceleration during the time is $2.09\times {10}^3{\text{m/s}}^{\text{2}}$

Explanation of Solution

Givendata:

Distance, for take-off speed $\Delta y=0.428\text{m}$

Distance for acceleration necessary to attain the required speed from rest $\Delta y=0.00200\text{m}$

Final speed $v_y=2.89\text{m/s}$

Initial speed $v_{0y}=0$

  $m_{\text{froghopper}}=12\text{mg}\text{=}12.3\times {10}^{-3}\text{kg}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $a_y=\frac{\Delta v_y}{\Delta \mathrm{t}}$

  $v^2=u^2+2a\Delta y$

Calculation:

We are interested in the portionof the jump while the insect is still on the ground.

We can assume that the acceleration of theinsect is constant during this phase

In order to calculate the acceleration of the insect and thelength of time during this phase of the jump, we first need to calculate the speed with whichthe froghopper leaves the ground.

While the insect is in the air, it is only under the influenceof gravity, so we can use the constant acceleration equations and the height of the jump tocalculate the takeoff speed.

Once we have this value, we know the froghopper acceleratedfrom rest through a distance of $2.00\text{mm}$to reach this takeoff speed. We can directly calculatethe time because we are assuming the acceleration is constant.

The forces acting on the insectwhile it is on the ground are the force of the ground on the froghopper pointing up (that is,the normal force) and the force of gravity pointing down.

After defining a coordinate systemwhere positive y points upward, we can calculate the magnitude of the normal force usingNewton's second law.

Takeoff speed is,

  $\begin{array}{c}{v}_y{}^2=v^2{}_{0y}+2a_y(y-y_0)\\ v_{0y}=\sqrt{v_y{}^2-2a_y(\Delta y)}=\sqrt{0-2\left(-9.80{\text{m/s}}^{\text{2}}\right)\left(0.428\text{m}\right)}\\ =2.89\text{m/s}\end{array}$

Acceleration necessary to attain that speed from rest is,

  $\begin{array}{l}{v}_y{}^2=v^2{}_{0y}+2a_y(y-y_0)\\ \Rightarrow a_y=\frac{v_y{}^2-v^2{}_{0y}}{2(\Delta y)}=\frac{{(2.89\text{m/s})}^2-0}{2(0.00200\text{m})}=2.09\times {10}^3{\text{m/s}}^{\text{2}}\end{array}$

Time is,

  $\begin{array}{l}{a}_y=\frac{\Delta v_y}{\Delta t}\\ \Delta t=\frac{(\Delta v_y)}{a_y}=\frac{(2.89\text{m/s})}{2.09\times {10}^3{\text{m/s}}^{\text{2}}}=0.00138\text{s}=1.38\text{ms}\end{array}$

Conclusion:

The time of the froghopper jump is $1.38\text{ms}$ and the acceleration during the jump is calculated as $2.09\times {10}^3{\text{m/s}}^{\text{2}}$.

To determine

(b)

The free body diagram of the grasshopper during the leap.

Explanation of Solution

Free body diagram of the grasshopper during the leap is,

  

Where, ${\overrightarrow{w}}_{\text{froghopper}}$ represents the effect of component of weight of froghopper and

  $\overrightarrow{n}$ represents the normal component which is always opposite to the effect of component of weight.

To determine

(c)

The force that ground exert on the froghopper during the jump in millinewtons and represent it as a multiple of insect's weight.

Answer to Problem 91QAP

The force that ground exert on the froghopper during the jump in millinewtons is

And the force $214$times larger than the froghopper's weight.

Explanation of Solution

Given data:

Distance, for take-off speed $\Delta y=0.428\text{m}$

Distance for acceleration necessary to attain the required speed from rest $\Delta y=0.00200\text{m}$

Final speed $v_y=2.89\text{m/s}$

Initial speed $v_{0y}=0$

  $m_{\text{froghopper}}=12\text{mg}\text{=}12.3\times {10}^{-3}\text{kg}$

Formula Used:

Newton's second law:

  $F_{\text{net}}=m\times a$

  $a_y=\frac{\Delta v_y}{\Delta \mathrm{t}}$

  $v^2=u^2+2a\Delta y$

Calculation:

We are interested in the portionof the jump while the insect is still on the ground.

We can assume that the acceleration of the insect is constant during this phase

In order to calculate the acceleration of the insect and the length of time during this phase of the jump. We first need to calculate the speed with which the froghopper leaves the ground.

While the insect is in the air, it is only under the influence of gravity, so we can use the constant acceleration equations and the height of the jump to calculate the takeoff speed.

Once we have this value, we know the froghopper accelerated from rest through a distance of $2.00\text{mm}$to reach this takeoff speed.

We can directly calculate the time because we are assuming the acceleration is constant.

The forces acting on the insect while it is on the ground are the force of the ground on the froghopper pointing up (that is, the normal force) and the force of gravity pointing down.

After defining a coordinate system where positive y points upward, we can calculate the magnitude of the normal force using Newton's second law.

  $\begin{array}{l}\sum F_{\text{ext,y}}=n-w_{\text{froghopper}}=n-m_{\text{froghopper}}g=m_{\text{froghopper}}{a}_y\\ n=m_{\text{froghopper}}g+m_{\text{froghopper}}{a}_y=m_{\text{froghopper}}(g+a_y)\\ n=\left(12.3\times {10}^{-3}\text{kg}\right)\left(\left(9.80{\text{m/s}}^{\text{2}}\right)+\left(2.09\times {10}^3{\text{m/s}}^{\text{2}}\right)\right)\\ n\text{=25}\text{.8}\text{N=0}\text{.0258}\text{mN}\end{array}$

froghopper's weight $=12.3\times 9.8=120.54\text{N}$

  $\begin{array}{l}\frac{n}{w_{\text{froghopper}}}\text{=}\frac{\text{25}\text{.8}\text{N}}{120.54\times {10}^{-3}\text{N}}=214\\ n=214w_{\text{froghopper}}\end{array}$

This force is $214$ times larger than the froghopper's weight.

Conclusion:

Thus, we have the value of force $\text{=0}\text{.0258}\text{mN}$

And the force is $214$ times larger than the froghopper's weight.