Question

a. How much work must you do to push a $10$kg block of steel across a steel table at a steady speed of $1.0$m/s for $3.0$s?

b. What is your power output while doing so?

Short answer

a). work done is $176.4$J

b). power output is$58.8$W

Step-by-step solution

Step 1. Part a). Find the work done

Work done on block is $W=F∆r$, where we don't know (yet) force or displacement. If we look at force on block, we see that there are two on block: friction and force of push. We can use second law of Newton

$$\begin{gathered} m\overrightarrow{a}=\sum _{}\overrightarrow{F} \\ 0=F_{friction}-F_{push} \\ {F}_{push}=mg{\mu }_{steeltosteel} \\ =10kg·9.8m/s^2·0.6 \\ =58.8 \end{gathered}$$

Displacement with velocity and elapsed time

$∆r=v∆t=1m/s·3=3m$

So, work required to push the block is

$$\begin{gathered} W=F∆r \\ =58.8N·3m \\ =176.4J \end{gathered}$$

Step 2. Part b). Power Output

$$\begin{gathered} P=\frac{∆W}{∆t} \\ P=\frac{176.4}{3} \\ P=58.8 \end{gathered}$$