Question

the coefficient of kinetic friction between a 20 kg box and the floor is 0.4. How much work does a pulling force do on the box in pulling it 8 m across the floor at constant speed? The pulling force is directed 37 degree above the horizontal.

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Coefficient}\mathrm{of}\mathrm{kinetic}\mathrm{friction},\mathrm{\mu }=0.4 \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{box}=20\mathrm{kg} \\ \mathrm{Distance},\mathrm{x}=8\mathrm{m} \\ \mathrm{The}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{force}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{W}=\mathrm{Fs}\mathrm{cos\theta } \\ =\mathrm{Fxcos}{37}^0 \\ \mathrm{Where},\mathrm{Fcos}{37}^0=\mathrm{f}=\mathrm{\mu }{\mathrm{F}}_{\mathrm{N}} \\ =\mathrm{\mu }[\mathrm{mg}-\mathrm{Fsin}{37}^0] \\ \mathrm{Or}\mathrm{F}=\frac{\mathrm{\mu mg}}{[\cos {37}^0+\mathrm{\mu sin}{37}^0]} \\ =\frac{0.4\times 20\times 9.8}{[\cos {37}^0+0.4\times \sin {37}^0]} \\ =75.4\mathrm{N} \\ \mathrm{Now}, \\ \mathrm{W}=75.4\cos {37}^0/8 \\ =482\mathrm{J} \end{gathered}$$