The efficiency of the machine is defined as
[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]
Here
Work out is the work output and Work in is the work input
To find the Work in we have then
[tex]W_{in} = \frac{W_{out}}{\eta}[/tex]
[tex]W_{in} = \frac{mgh}{\eta}[/tex]
Replacing with our values
[tex]W_{in} = \frac{(58)(9.8)(3)}{73\%}[/tex]
[tex]W_{in} = 2335.89J[/tex]
The work done by the applied force is
W = Fd
Here,
F = Force
d = Distnace
Rearranging to find F,
[tex]F = \frac{W}{d}[/tex]
[tex]F = \frac{2335.89J }{18}[/tex]
F = 129.77N
Therefore the force exerted on the machine after rounding off to two significant figures is 130N
