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?

Answer 1

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. $N=14.7W.$