Answer:
A. Input work = 800 ft.lb
B. Output work = 750 ft.lb
C. Efficiency of block and tackle = 93.75 %
Explanation:
Given:
Weight of the load, [tex]W=600[/tex] lb.
Distance moved by the load, [tex]D_{load}=1.25[/tex] ft.
Force applied on the rope, [tex]F = 200[/tex] lb
Distance moved by the force on the rope, [tex]D{in}=4[/tex] ft.
Work done by a force causing a displacement in the direction of force is given as:
[tex]\textrm{Work}=\textrm{Force}\times \textrm{Displacement}[/tex]
A.
Here, Input Work is given by the input force and the displacement caused by the input force. So,
[tex]W_{in}=F\times D_{in}\\W_{in}=200\times 4=800\textrm{ ft.lb}[/tex]
Therefore, the input work is 800 lb.ft
B.
Output Work is given by the output force and the displacement caused by the output force. So,
[tex]W_{out}=F_{load}\times D_{load}\\W_{out}=600\times 1.25=750\textrm{ ft.lb}[/tex]
Therefore, the output work is 750 ft.lb
C.
Efficiency is given as the ratio of Output Work to Input Work expressed as percentage. So,
Efficiency = [tex]\frac{W_{out}}{W_{in}}\times 100=\frac{750}{800}\times 100=93.75\%[/tex]
Therefore, efficiency of block and tackle is 93.75 %.
Answer:
A. Input work = 800 ft.lb
B. Output work = 750 ft.lb
C. Efficiency of block and tackle = 93.75 %
