Question

A $45$g bug is hovering in the air. A gust of wind exerts a force role="math" localid="1648061852575" $\overrightarrow{F}=(4.0\hat{i}-6.0\hat{j})\times {10}^{-2}$N on the bug.

a. How much work is done by the wind as the bug undergoes displacement $∆\overrightarrow{r}=(2.0\hat{i}-2.0\hat{j})$m?

b. What is the bug’s speed at the end of this displacement? Assume that the speed is due entirely to the wind.

Short answer

a). Work done is $0.20$J

b). speed $v_f=3$m/s

Step-by-step solution

Step 1. Given Information

Mass $m=45$g

Exert force$\overrightarrow{F}=(4.0\hat{i}-6.0\hat{j})\times {10}^{-2}$

Step 2. Part a). Find the work done

A force $F$that acted on an object with mass $m$is parallel to the direction of motion and causes displacement $∆r$for the object.

$W=\overrightarrow{F}∆\overrightarrow{r}$

The dot product of the two vectors $\overrightarrow{F}$and $∆\overrightarrow{r}$. So,

$W=F_x∆r_x+F_y∆r_y$

Now,

$$\begin{gathered} W=F_x∆r_x+F_y∆r_y \\ =(4\times {10}^{-2}N)(2m)+(-6\times {10}^{-2}N)(-2m) \\ =0.20J \end{gathered}$$

Step 3. Part b). Find the speed

The change in the kinetic energy$∆K$represents the work done$W$

So,

$W=K_f-K_i$

The work done by the tension in the form

$$\begin{gathered} W=K_f-K_i \\ W=\frac{1}{2}m{v^2}_f-\frac{1}{2}m{v^2}_i \\ W=\frac{1}{2}m\left[{v^2}_f-{v^2}_i\right] \\ {v^2}_f=\frac{2W}{m}+{v_i}^2 \\ {v}_f=\sqrt{\frac{2W}{m}+{v_i}^2} \end{gathered}$$

The initial speed of the bug is zero

$$\begin{gathered} v_f=\sqrt{\frac{2(0.20)}{45\times {10}^{-3}}+{\left(0\right)}^2} \\ =3m/s \end{gathered}$$