Question

the mechanical advantage of a machine is 5 and its efficiency is 80%. It is used to lift a load of 200 kgf to a height of 20m. Calculate :
i) the effort required
ii)the work done on the machine (g=10 ms^-1)

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{mechanical}\mathrm{advantage},\mathrm{MA}=5 \\ \mathrm{efficiency},\mathrm{\eta }=80\% \\ \mathrm{load},\mathrm{W}=200\mathrm{kgf} \\ \mathrm{or}\mathrm{W}=200\times 10\mathrm{N} \\ \mathrm{W}=2000\mathrm{N} \\ \mathrm{distance}\mathrm{moved}\mathrm{by}\mathrm{load},\mathrm{y}=20\mathrm{m} \\ \mathrm{Effort},\mathrm{P}=\frac{\mathrm{W}}{\mathrm{MA}} \\ \mathrm{P}=\frac{2000}{5}=400\mathrm{N} \\ \mathrm{Velocity}\mathrm{ratio},\mathrm{VR}=\frac{\mathrm{MA}}{\mathrm{\eta }} \\ \mathrm{VR}=\frac{5}{{\displaystyle \frac{80}{100}}} \\ \mathrm{VR}=6.25 \\ \mathrm{Also},\mathrm{VR}=\frac{\mathrm{distance}\mathrm{moved}\mathrm{by}\mathrm{effort}\left(\mathrm{x}\right)}{\mathrm{distance}\mathrm{moved}\mathrm{by}\mathrm{load}\left(\mathrm{y}\right)} \\ 6.25=\frac{\mathrm{x}}{20} \\ \mathrm{x}=125\mathrm{m} \\ \mathrm{Work}\mathrm{done}\mathrm{on}\mathrm{machine},\mathrm{W}\text{'}=\mathrm{P}\times \mathrm{x} \\ \mathrm{W}\text{'}=400\times 125=50000\mathrm{J} \end{gathered}$$