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

b. What is your power output while doing so?

Short Answer

Expert verified

a). work done is 176.4J

b). power output is58.8W

Step by step solution

01

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

ma→=∑F→0=Ffriction-FpushFpush=mgμsteeltosteel=10kg·9.8m/s2·0.6=58.8

Displacement with velocity and elapsed time

∆r=v∆t=1m/s·3=3m

So, work required to push the block is

W=F∆r=58.8N·3m=176.4J

02

Step 2. Part b). Power Output

P=∆W∆tP=176.43P=58.8

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