a) 53.0 cm
b) 67.7 cm
Explanation:
a)
In an ideal lever (100% efficiency), the work done in input is equal to the work done in output. So we can write:
[tex]W_{in} = W_{out}[/tex]
And both works can be rewritten as
[tex]F_i d_i = F_o d_o[/tex]
where
[tex]F_i[/tex] is the force in input
[tex]F_o[/tex] is the force in output
[tex]d_i[/tex] is the effort arm
[tex]d_o[/tex] is the load arm
Here we have:
[tex]F_i=218 N[/tex] is the force exerted in input
[tex]F_o=1050 N[/tex] is the output force (the weight of the rock)
[tex]d_o = 11.0 cm[/tex] is the distance through which the rock is lifted
So, we can find the distance through which the lever moves on the input end:
[tex]d_i = \frac{F_o d_o}{F_i}=\frac{(1050)(11.0)}{218}=53.0 cm[/tex]
b)
In this case, the lever has an efficiency of
[tex]\eta = 78.3\% = 0.783[/tex]
Efficiency can be rewritten as the ratio between output work and input work:
[tex]\eta=\frac{W_o}{W_i}[/tex]
The output work is
[tex]W_o=F_o d_o =(1050)(11)=11,550 N\cdot cm[/tex]
Therefore the input work is
[tex]W_i = \frac{W_o}{\eta}=\frac{11550}{0.783}=14,751 N\cdot cm[/tex]
This input work can be rewritten as
[tex]W_i = F_i d_i[/tex]
And so we can find by how much the girl moves her end of the lever in this case:
[tex]d_i=\frac{W_i}{F_i}=\frac{14751}{218}=67.7 cm[/tex]
