Solution.
"m=10kg;"
"\\upsilon=1m\/s;"
"t=3.0s;"
"\\mu=0.15" - sliding coefficient of steel steel;
a. "A=Fl;"
Since the unit moves evenly, the traction force is equal to the friction force:
"F=\\mu N; N=mg; F=\\mu mg;"
"F=0.15\\sdot10kg\\sdot9.8N\/kg=14.7N;"
"l=\\upsilon t; l=1m\/s\\sdot3.0s=3.0m;"
"A=14.7N\\sdot3.0m=44.1J;"
b. "N=\\dfrac{A}{t}" - the power you develop;
"N=\\dfrac{44.1J}{3.0s}=14.7W;"
Answer:a. "A=44.1J;"
b. "N=14.7W."
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