Answer:
Approximately [tex]0.45[/tex] (assuming that the surface is level, the mass is moving at constant velocity, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
With a power of [tex]P = 1.0\; {\rm kW} = 1.0\times 10^{3} \; {\rm W}[/tex], the work that would be done over [tex]t = 10\; {\rm s}[/tex] would be:
[tex]\begin{aligned}(\text{work}) &= (\text{power})\, (\text{time}) \\ &= (1.0\times 10^{3}\; {\rm W})\, (10\; {\rm s}) \\ &= 1.0 \times 10^{4}\; {\rm J}\end{aligned}[/tex].
Divide work by distance to find the force that did the work:
[tex]\begin{aligned} (\text{force}) &= \frac{(\text{work})}{(\text{distance})} \\ &= \frac{1.0\times 10^{4}\; {\rm J}}{50\; {\rm m}} \\ &= 200\; {\rm N}\end{aligned}[/tex].
If this mass is moving at a constant velocity, the magnitude of friction on this mass will be equal to that of the external force, [tex]200\; {\rm N}[/tex].
If the surface is level, the magnitude of the normal force on this mass will be equal to that of weight:
[tex]\begin{aligned}(\text{weight}) &= (\text{mass})\, g \\ &= (45\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-2}}) \\ &\approx 4.41 \times 10^{3}\; {\rm N}\end{aligned}[/tex].
Divide the magnitude of friction by normal force to find the coefficient of friction:
[tex]\begin{aligned}& (\text{coefficient of friction}) \\ =\; & \frac{(\text{friction})}{(\text{normal force})} \\ \approx\; & \frac{200\; {\rm N}}{(45\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}))} \\ \approx \; & 0.45\end{aligned}[/tex].
