Question

A $25$kg air compressor is dragged up a rough incline from $\overrightarrow{r_1}=(1.3\hat{i}+1.3\hat{j})$m to $\overrightarrow{r_2}=(8.3\hat{i}+2.9\hat{j})$m, where the$y-$axis is vertical. How much work does gravity do on the compressor during this displacement?

Short answer

Work done by gravity is$392$J

Step-by-step solution

Step 1. Given Information

Mass of the air compressor$=25$Kg

Initial Position $r_1=1.3\hat{i}+1.3\hat{j}$

Final position$r_2=8.3\hat{i}+2.9\hat{j}$

Step 2. Find the work done

We have to determine the work done by gravity on the air compressor

It is given that air compressor is being dragged up ab incline, hence the displacement is against gravity in opposite direction to gravity force. Gravity acts downward and taking downward direction as $y-$axis we will calculate the work done

$$\begin{gathered} \overrightarrow{F_G}=-mg\hat{j} \\ =-25\times 9.8 \\ =-245N\hat{j} \\ \overrightarrow{∆r}=\overrightarrow{r_2}-\overrightarrow{r_1} \\ =(-245\hat{j})·(7.0\hat{i}+1.6\hat{j}) \\ W=\overrightarrow{F}·\overrightarrow{∆r} \\ =(-245\hat{j})·(7.0\hat{i}+1.6\hat{j}) \\ =-245\times 7\left(\hat{j}·\hat{i}\right)-245\times 1.6(\hat{j}·\hat{i}) \\ =-245\times 7\times \cos 90°-245\times 1.6\times \cos 0° \\ =0-392J \end{gathered}$$

The negative sign of work indicates that the work done by gravity is against the force applied to drag air compressor up